Precalculus OLD
This course fills the minimum mathematics course requirement for students who plan to participate in postsecondary education. It also serves as the prerequisite for Advanced Placement Calculus or Statistics courses. In preparation for this course, students should have mastered linear and quadratic functions, concepts from discrete mathematics involving sequences and series, and data analysis and probability techniques. Students should also be able to confidently work with expressions containing rational exponents and radical and rational terms.
The first problem type we will consider is how to solve a triangle, given the length of one side, and the measure of an angle. We will use both the definitions of the trigonometric functions, and our calculator.
Consider the triangle shown here. If A = 35° and c = 16, find the remaining sides and angles.

Okay, to solve this problem, we will use both the sine and the cosine.
First, since side a is opposite angle A, the sine of A is
(eq. 5)
Solve this for ,
= c sin A (eq. 14)
and plug in the numbers
= 16 sin 35° = (16)(0.5735...) = 9.178.... (eq. 15)
Now we need to find the length of side b. Side b is adjacent to angle A, so the cosine of A is
(eq. 6)
Solve this for b,
b = c cos A (eq. 16)
and plug in the numbers
b = 16 cos 35° = (16)(0.8191...) = 13.106.... (eq. 17)
Next, we need to find the measure of angle B. We know that
A + B = 90° (eq. 13)
Therefore,
B = 90° – A (eq. 18)
Again, plug in the numbers
B = 90° – 35° = 55° (eq. 19)
This is everything. The triangle is defined by the equations:
A = 35° (eq. 20)
B = 55° (eq. 19)
C = 90° (eq. 21)
= 9.178 (eq. 15)
b = 13.106 (eq. 17)
c = 16 (eq. 22)
Easy enough? What would you do differently if you were given angle B instead?
When we say that we are going to “solve a triangle,” it means that given enough information, we can find all the lengths of the sides, and all the measures of the angles for an arbitrary triangle. The simplest triangle type to solve are the right triangles. Since the trigonometric functions are defined by the right triangles, this is logical. Consider the following right triangle.
It is standard to name triangles by their vertices, and vertices are named with capital letters. The triangle here is named ΔABC. While triangles can be named anything really, this is the name we see
most often.
For a right triangle, usually we name the vertex with the right angle C, and the other vertices A and B. It is standard to name the angle at the vertex by the same name as the vertex. So, the triangle shown has vertices A, B, and C, with angles A, B and C located at those vertices, respectively.

It is also standard to name the side opposite the angle with the same name as the angle, but lowercase. So the side opposite angle A is side a, the side opposite angle B is side b, and the side opposite angle C is side c.
Finally, if we italicize the name, A, B, and C for angles or , b, and c for sides, we are now discussing the measure of the angle, or the length of the side. Therefore: angle A has measure A, angle B has measure B, angle C has measure C, side a has length , side b has length b, and side c has length c.
Using this naming convention, the Pythagorean theorem is
c^{2} = ^{2} + b^{2} (eq.1)
Because of the Pythagorean theorem, if we know any two sides of a right triangle, we can solve for the third side.
Also, recall, the trigonometric functions are defined by right triangles. In particular
(eq. 2)
(eq. 3)
and
(eq. 4)
For our right triangle, this means that
(eq. 5)
(eq. 6)
(eq. 7)
and
(eq. 8)
(eq. 9)
(eq. 10)
Using these definitions, if we know one side and one angle, we can find the other side. Or if we know two sides, we can use the inverse trigonometric functions to find the angle.
Finally, we know that the sum of all the angles is 180°, and that the right angle has a measure of 90°. Therefore, if we know one angle, we really know all three. In particular
A + B + C = 180° (eq. 11)
which is true for all triangles. For right triangles we also have
A + B + 90° = 180° (eq. 12)
or
A + B = 90° (eq. )
Using these relations, if we know two sides or one side and one angle, we can solve any right triangle.
For most of this quarter, we have tried to get away form the notion of trigonometric functions as being properties of triangles, and instead have been focusing on trigonometric functions as functions. However, the trigonometric functions are defined by triangles. It would be such a waste to not use these functions to “solve” triangles.
To derive the sumtoproduct identities, we simply make the change of variables
= A + B (eq. 8)
b = A – B (eq. 9)
which means that
(eq. 11)
em>(eq. 13)
Make those substitutions into the producttosum identities, and solve we find
(eq. 14)
(eq. 25)
(eq. 26)
(eq. 27)
Is that what you found also? These are the sumtoproduct identities. Make sure you have those written correctly as well.
Notice that each of these identities are different. It is somewhat surprising that we can take two identities that are the same and derive two identities that are different. The reason that this occurs has to do with the periodic and symmetric nature of these functions.
Now that we have the identities, we should solve a problem or four:
Example:
Rewrite the following sum of trigonometric functions as a product,
In the problem above, it doesn't matter which angle we let be and b. So, we might as well let _{} and _{}. Next, we want to find _{} and _{}.
(eq. 28)
(eq. 28a)
(eq. 28b)
(eq. 28c)
(eq. 29)
(eq. 29a)
(eq. 29b)
(eq. 29c)
(eq. 29d)
I went through each step above just as a “how to add fractions” review. You should have been able to follow that and correctly add other fractions as needed.
Okay, so we have that _{} and _{}. We can now enter these terms into equation 14.
(eq. 30)
This isn't quite done. We need to use the cosine symmetry identity (equation 19) to simplify equation 30,
(eq. 31)
Let us repeat this for the other sums and differences of the trigonometric functions using the same values of and b as before.
Example:
Rewrite the following difference of trigonometric functions as a product,
We have already decided to let _{} and _{}, and we have found _{} and _{} . Therefore, we can just substitute these values into equation 25,
(eq. 32)
This time we want to use the sine symmetry identity (equation 23) to simplify this,
(eq. 33)
... and that is it! Next problem.
Example:
Rewrite the following sum of trigonometric functions as a product,
We already have identified and calculated all the angles we need. Just substitute those into equation 26,
(eq. 34)
Use the symmetry identity to simplify,
(eq. 35)
Ready to do the last problem?
Example:
Rewrite the following difference of trigonometric functions as a product,
We know everything, just substitute into equation 27,
(eq. 36)
Use the symmetry identity to simplify,
(eq. 37)
Tuhduh! We are done.
Ready to do this on your own?
We derived the first of the producttosum identities together. You derived the remaining three on your own. So, just so you can check your results, the identities are
(eq. 4)
(eq. 15)
(eq. 16)
(eq. 17)
Did you get the correct answers? Awesome! Make sure you have these written correctly in your notes.
Clearly, equations 16 and 17 are different. You can use equation 16 if you have something like
cos 73° cos 108°
and you wish to write this as a sum instead. (Can you do that? Take a second and try it.)
Also, if you have
sin 218° sin 6°
and you wish to write that as a sum, the only equation that will work is equation 17. But what if you have
sin 98° cos 239° ?
Is that really different from
cos 239° sin 98°?
No, they are the same. This is one of the fundamental properties of multiplication. Right? So, if you want to write this expression as a sum, you should be able to use either equation 4 or equation 15. How do you know which one to use?
They are exactly the same. It doesn't matter. Can you prove that these equations are the same? They don't look the same. That is a good test to see if you remember the basic trigonometric identities from lesson 1. (That is a hint.)
Since we have a set of problems, let's go ahead and practice these.
Example: Rewrite the following product of trigonometric functions as a sum or difference,
cos 73° cos 108°
To do this, we want to use equation 16. Let A = 73° and B = 108°,
(eq. 18)
(eq. 18a)
Technically, this is okay, but it is bad form to leave a negative angle inside of the cosine, so let's just simplify that. You should recall that
cos(x) = cos x (eq. 19)
this is the symmetry identity for the cosine. Therefore, equation 18a can be simplified to
(eq. 20)
This is the same result we would have found if we had let A = 108° and B = 73°, so we are good.
Example: Rewrite the following product of trigonometric functions as a sum or difference,
sin 218° sin 6°
To write this as a sum or difference we need to use equation 17. Let A = 218° and B = 6°. This gives us,
(eq. 21)
(eq. 21a)
Can you demonstrate that this would be the same if we let A = 6° and B = 218°? (Hint: use equation 19 again.)
Example: Rewrite the following product of trigonometric functions as a sum or difference,
cos 239° sin 98°
Technically, we could use either equation 4 or equation 15. However, equation 15 is already in the same order, so use that one. Let A = 239° and B = 98°. This gives us,
(eq. 22)
(eq. 22a)
Just as it was bad form to have 35° inside the cosine in equation 18a, it is bad form to leave 141° in the sine in equation 22a. Do we know a symmetry identity for sines? Hint:
sin(x) = sin x (eq. 23)
Use that identity to simplify equation 22a,
(eq. 24)
Does that look more like what you expected from equation 4? Try it.
While you are working on deriving the full set of producttosum identities, let's consider the sumto product identities. Without being very clever, you could argue that we already have these identities. If you take equation 3 and do nothing, isn't that already a “sumtoproduct” identity?
Yes and no. For that identity to be useful, we really need the terms on the lefthand side to be single variables, and anything else we want to do with the variables should be done on the left. In other words, instead of having
sin(A + B) + sin(A – B) = 2 sin A cos B (eq. 3a)
we really want something more like
sin(C) + sin(D) = 2 sin(ƒ(C,D)) cos((C,D)) (eq. 7)
In equation 7, C and D are arbitrary angles. ƒ(C,D) and (C,D) are unknown functions of the angles C and D. Okay? So, why don't we just solve that problem?
What we really want to do here is change the variables. We know that we want A + B to be a single angle, not a sum of angles. Our goal is not to have to solve a system of equations every time we want to convert a sum of trigonometric functions to a product of trigonometric functions. It is to solve that problem from the beginning, then we can just use the new identity whenever we need it.
Therefore, we will make a change of variables. In particular, we will let
= A + B (eq. 8)
and
b = A – B (eq. 9)
Now, we will solve these equations for A and B. By this point in your math career you probably have a “favorite” method for solving a system of linear equations. Go ahead and use whichever method you like the best. I like to add and subtract the equations. As we did before,
Divide both sides of equation 10 by 2, and reorder,
If we repeat this, but subtract equation 9 from equation 8 instead, we get,
Alright? Now, we just substitute equations 11 and 13 for A and B, and equations 8 and 9 for A + B and A – B. If we do this, equation 3 becomes
Cool? So that is the first of the sumtoproduct identities.
By making the same substitution to the producttosum identities you derived, you can generate the remaining sumtoproduct identities.
So far in this class: You have simplified both simple and complex trigonometric expressions. You have completed dozens of proofs. You have solved fairly complicated trigonometric problems. You have proven both your algebraic and your trigonometric skills. You are ready to take it one step further. You are ready to derive for yourself these identities.
In the same way that we were able to derive the double angles from the sum identities, we can derive the product and sum identities from the sum and difference identities.
The following equations are the sine sum and difference identities,
sin(A + B) = sin A cos B + cos A sin B (eq. 1)
sin(A – B) = sin A cos B – cos A sin B (eq. 2)
If we add these equations together, we get
Next, divide both sides of this equation by 2, and reorder the equation. We will find
This is the first of the producttosum identities. Write it down.
So, what will happen if you subtract equation 2 from equation 1? Take a minute and do this. What do you find? Seriously, do this, you can check your answer in section 3.
That is about all you can do with the sine sum and difference identities. Now, consider the cosine sum and difference identities.
cos(A + B) = cos A cos B – sin A sin B (eq. 5)
cos(A – B) = cos A cos B + sin A sin B (eq. 6)
Can you see that if we add and subtract these we will be able to generate 2 additional producttosum identities. Again, take a minute and do this yourself. We will discuss these results later.
￼
What did you find? How are these identities different from the ones you found using the sine sum and difference identities? Did you notice anything weird about the producttosum identities you found with the sine sum and difference identities?
From the names, you will probably not be surprised that the product and sum identities take a the sum of two trigonometric functions and convert these to a product, or take the product of two trigonometric functions and convert these to a sum. There will be times in your mathematical career when you may have a sum, but need a product, or have a product but need a sum.
In the previous lesson, we decided to investigate the problem of the launch angle of a projectile that travels through the air such as a cannon, catapult, or a popular video game about birds. Cannons, catapults, and birds launched at pigs are projectiles that behave almost like ideal projectiles. The equations that describe the motion of a projectile are
x = v cos θ t (eq. 78)
(eq. 79)
Where x is the horizontal distance, y is the vertical distance, v is the initial speed, θ is the launch angle, t is the time, and _{} is the acceleration of gravity, which on the Earth is a constant equal to 9.8 m/s or 32 ft/s depending on the units. In a real battle, there isn't much we can do about the initial speed, what we can change is the launch angle. So assume you know how far away your target is, the trigonometry problem is what angle should we use to hit a target at a known location.
Okay? Let's look at the knowns and unknowns in this problem: x is the horizontal distance. This is the location of our enemy. For this problem, we will assume that we know this distance (or can estimate it fairly accurately). v is the initial speed of our projectile. Again, this is something that we would know in advance, we measured this at the test range long before the battle began. t is the time. This is a variable. In particular, it is the independent variable, and we cannot do anything to change it. θ is the angle we wish to launch the projectile in order to hit the enemy. This is the variable we wish to solve for, since we can control the launch angle. Therefore it is the dependent variable. Finally y is the vertical distance. If we are at the same level as the enemy this will be 0, ie, we are on relatively flat ground, not on a hill or something. We already mentioned that g is the acceleration of gravity on the Earth, and it is a constant.
Alright, still with me? x, v, y and _{} are constants, and y = 0. θ and t are variables. We want to find θ, we don't really care much about t. Actually, let me correct that thought. Imagine we can fire a cannon bomb instead of just a cannon ball. Then we may have some ability to change the length of the fuse so that the bomb explodes just before it lands. That would be good. So, we may wish to solve for t as well as θ.
If we let y = 0, equation 79 becomes
(eq. 80)
Which we can now simplify,
(eq. 81)
So far so good? Next since we really want to find θ, we should eliminate t. First, use equation 78 to solve for t,
(eq. 82)
Plug equation 82 into equation 81,
(eq. 83)
Simplify,
(eq. 83)
Isolate θ,
(eq. 84)
(eq. )
Well, sort of isolate θ. Notice that we multiplied both sides by 2 cos θ, and not just cos θ. We did this for a reason. You should recognize 2 sin θ cos θ as a double angle identity. In particular,
sin(2 θ) = 2 sin θ cos θ (eq. 40c)
Make that substitution,
(eq. 86)
Of course, you know how to solve it from here. Take the inverse sine, divide that value by 2.
Just in case you do want to solve for the time, plug that answer into equation 82. No, this isn't very pretty, but it was sort of a fun journey.
This is a very real transcendental problem that you know how to solve. Congrats.
Obviously, we can't finish the lesson without an example. Let's use the triangle from section 5. We already determined that the sine of θ and the cosine of θ are:
(eq. 52)
(eq. 53)
Therefore, we can find the three basic trigonometry functions for by using equation 64, 68, and 70 or 71a or 72a. Start with equation 64.
(eq. 64)
θ is less than 90°, therefore is also less than 90°. This means that we use the positive root in equation 64. Start by substituting equation 53 in equation 64. Oh, and let B = θ,
(eq. 73)
... and simplify
(eq. 73a)
(eq. 73b)
(eq. 73c)
(eq. 73d)
Repeat for the cosine and tangent. This time, I will not go through each of the steps to simplify these. You should be able to simplify an expression. Start with equation 68 to find the cosine of . Again, ￼take the positive root, since it is in the first quadrant.
(eq. 68)
(eq. 74)
(eq. 74a)
Next, let's find all three of the halfangle tangent identities. Start with equation 70. First quadrant, so positive root.
(eq. 70)
(eq. 75)
(eq. 75a)
(eq. 75b)
... and just to double check, equations 71a and 72a.
(eq. 71a)
(eq. 76)
(eq. 76a)
(eq. 75b)
(eq. 72a)
(eq. 77)
(eq. 77a)
(eq. 75b)
Three equations, one answer. How cool is that?
So, do you think you are ready to do this on your own?
The last set of identities we will be finding in this lesson are the halfangle identities. These identities can be derived from the doubleangle identities. In particular, from the cosine doubleangle identities. Consider the third of the cosine doubleangle identities,
cos(2 A) = 1 – 2 sin^{2} A (eq. 46a)
Let's do a change of variables. Let ,
(eq. 61)
(eq. 61a)
Now, solve this equation for sin .
(eq. 62)
(eq. 63)
(eq. 64)
We use the “±” sign because the actual value depends on the location of . If is in the first or second quadrant, keep the positive root. If is in the third or fourth quadrant, keep the negative root. Oh, and just in case you didn't realize, equation 64 is the first of the halfangle identities. Write it down.
We can use the second of the cosine doubleangles to find the second halfangle identity.
cos(2 A) = 2 cos^{2} A – 1 (eq. 44b)
Make the same substitution as before,
(eq. 65)
(eq. 65a)
Solve for
(eq. 66)
(eq. 67)
(eq. 68)
Equation 68 is the second of the halfangle identities. Write it down also. Again, the “±” sign depends on the location of . If is in the first or fourth quadrant, keep the positive root. If is in the second or third quadrant, keep the negative root.
There are three different halfangle tangent identities. The first can be found by dividing equation 64 by equation 68,
(eq. 69)
(eq. 70)
Equation 70 is the first of the tangent halfangle identities. Write it down. The “±” sign depends on the location of . If is in the first or third quadrant, keep the positive root. If is in the second or fourth quadrant, keep the negative root.
Personally, I find this to be the least useful of the tangent identities. Mostly because the other two do not have square roots. Recall, it is terrible form to have a square root in the denominator. We really should rationalize equation 70. (Hint.)
(eq. 70a)
(eq. 70b)
But recall, 1 – cos^{2} B is equal to sin^{2} B. Make that substitution,
(eq. 71)
(eq. 71a)
Equation 71a is the second halfangle tangent identity. Write it down.
Of course, in general we wouldn't want to “rationalize” the numerator. However, in this case, it works out nicely. So, let's try it,
(eq. 70c)
(eq. 70d)
Again, substitute sin^{2} B for 1 – cos^{2} B in the denominator,
(eq. 72)
(eq. )
Equation 72a is the third halfangle tangent identity. Write it down as well.
Obviously, we should do an example. However, instead of using the special angles, let's use a known triangle. Consider the triangle below:

We do not know the value of θ, but we do know (or can calculate) all the sides. Therefore, we can determine all the trigonometric functions for θ. Once we know these, we can calculate the basic trigonometric functions for 2 θ using equation 40b, equation 41a or 44b or 46a and equation 48.
First, let's calculate the hypotenuse. We do this with the Pythagorean theorem;
hyp^{2} = opp^{2} + adj^{2} (eq. 49)
which gives us
hyp^{2} = 7^{2} + 24^{2} = 625 (eq. 50)
hyp = 25 (eq. 51)
Therefore, the three basic trigonometric functions of θ are:
(eq. 52)
(eq. 53)
(eq. 54)
Now we can use these, as well as the aforementioned equations, to find the three basic trigonometric functions of 2 θ. Start with equation 40b,
sin(2 A) = 2 sin A cos A (eq. 40b)
Substitute θ for A,
sin(2 θ) = 2 sin θ cos θ (eq. 40c)
Substitute equations 52 and 53 into equation 40c,
(eq. 55)
Multiply this out,
(eq. 55a)
Repeat for the cosine of 2 θ and the tangent of 2 θ. I will use all three of the cosine equations, just to demonstrate that they result in the same answer. For your homework, you should be okay if you just use one of them. Although different problems may lend themselves better to one identity than another. Therefore, you really do need to be able to use all three.
￼
First cosine identity:
cos(2 θ) = cos2 θ – sin2 θ (eq. 41b)
(eq. 56)
(eq. 56a)
(eq. 56b)
Second cosine identity:
cos(2 θ) = 2 cos^{2} θ – 1 (eq. 44c)
(eq. 57)
(eq. 57a)
(eq. 56b)
Third cosine identity:
cos(2 θ) = 1 – 2 sin^{2} θ (eq. 46b)
(eq. 57)
(eq. 57a)
(eq. 56b)
Three identities, one answer. Coincidence? Nah. Finally, repeat with the tangent. I will not be going through every step in evaluating this.
(eq. 48a)
(eq. 58)
(eq. 58a)
(eq. 58b)
And for fun, notice that if we take sin 2 θ and divide this by cos 2 θ we get,
(eq. 59)
(eq. 60)
(eq. 58d)
Cool?
The next set of advanced identities that we will be considering are called the doubleangle identities. These are called this because they take the trigonometric functions of twice an angle, and write them in terms of the trigonometric functions of the angle.
We can derive each of the doubleangle identities from the sum identities. Again, while you do not need to memorize these, you will need to be able to access them while doing your homework. So, take a minute and write these down.
The first doubleangle identity is derived from the sine sum identity.
sin(A + B) = sin A cos B + cos A sin B. (eq. 1)
Start by letting B = A. Substitute that into equation 1,
sin(A + A) = sin A cos A + cos A sin A. (eq. 40)
Simplify,
sin(2 A) = sin A cos A + sin A cos A, (eq. 40a)
sin(2 A) = 2 sin A cos A. (eq. 40b)
And that is it. The first doubleangle identity is equation 40b. Write this down. To find the cosine doubleangle identity, use equation 3.
cos(A + B) = cos A cos B – sin A sin B, (eq. 3)
Let B = A, and substitute that into equation 3,
cos(A + A) = cos A cos A – sin A sin A, (eq. 41)
Simplify,
cos(2 A) = cos^{2} A – sin^{2} A. (eq. 41a)
Okay, equation 41a is the second doubleangle identity. Write it down.
However, we are not yet done. We can use the Pythagorean theorem to modify this. Recall, the basic Pythagorean theorem in trigonometry is
sin^{} A+cos^{2} A = 1. (eq. 42)
If we solve this for sin^{2} A we get
sin^{2} A = 1 – cos^{2} A. (eq. 43)
Substitute that back into equation 41a and we obtain,
cos(2 A) = cos^{2} A – (1 – cos^{2} A). (eq. 44)
This can be simplified,
cos(2 A) = cos^{2} A – 1 + cos^{2} A, (eq. 44a)
cos(2 A) = 2 cos^{2} A – 1. (eq. 44b)
Equation 44b is the next doubleangle identity. Write it down as well.
What if we had solved equation 42 for cos^{2} A instead? We would have found,
cos^{2} A = 1 – sin^{2} A. (eq. 45)
Substituting equation 45 into equation 41a gives us,
cos(2 A) = (1 – sin^{2} A) – sin^{2} A. (eq. 46)
This simplifies to,
cos(2 A) = 1 – 2 sin^{2} A. (eq. 46a)
Equation 46a is the last of the cosine doubleangle identities. Write it down.
Obviously, the tangent doubleangle identities will be found from the tangent sum identity,
(eq. 5)
Let B = A,
(eq. 47)
Simplify,
(eq. 48)
Equation 48 is the last of the doubleangle identities. Write it down.
Next we need to demonstrate the difference identities. We will continue with out original plan of using A = 45° and B = 30°, such that A – B = 15°.
Let's start by finding the sine of 15°. To do this, we use equation 2,
sin(AB) = sin A cos B – cos A sin B (eq. 2)
Plug in the values of A and B,
sin 15° = sin 45° cos 30° – cos 45° sin 30° (eq. 34)
Substitute the values of these trigonometric functions from equations 22, 23, 25 and 26.
(eq. 35)
Simplify this
(eq. 35a)
(eq. 35b)
Next use your calculator to evaluate the righthand side of equation 35b. You should find
(eq. 35c)
The lefthand side gives us
LHS = sin 15° = 0.2588190451.... (eq. 35d)
These are equal, so we have some evidence that equation 2 is valid. You can repeat this with as many more angles as you like if you want additional proof.
Next, find the cosine of 15°. Start with equation 4,
cos(A – B) = cos A cos B + sin A sin B (eq. 4)
Substitute the values of A and B,
cos 15° = cos 45° cos 30° + sin 45° sin 30° (eq. 36)
Substitute the values of these trigonometric functions,
(eq. 37)
Simplify,
(eq. 37a)
(eq. 37b)
Finally, use your calculator to evaluate both sides of this equation. You should find that
(eq. 37c)
LHS = cos 15° = 0.9659258263.... (eq. 37d)
These are equal. Therefore, equation 4 is possibly correct. Feel free to try a few more angles.
Finally, we wish to find the tangent of 15°. We need to use equation 6.
(eq. 6)
Plug in the values for A and B.
(eq. 38)
Then substitute the values from equations 24 and 27 into equation 38,
(eq. 39)
Simplify this
(eq. 39a)
(eq. 39b)
(eq. 39c)
Rationalize the denominator,
(eq. 39d)
(eq. 39e)
(eq. 39f)
(eq. 39g)
Now check your answer with your calculator,
(eq. 39h)
LHS = tan 15° = 0.2679491924.... (eq. 39i)
Tuh duh!
Of course, our previous derivation of equation 5 as an identity depends on equations 1 and 3 being identities. We did not prove equations 1 and 3. These identities (as well as equations 2 and 4) can be proven with geometry. We will not actually be going through this proof, but you can read it in the suggested links for this lesson. What we will do instead is use are calculator to look at a few examples to convince ourselves. This is not a proof per se, but it is a good way to disprove something.
We only know a dozen special angles. Let's pick the smallest, nonzero, pair of these to verify the sum and difference identities. In particular, consider 45° and 30°,
45° + 30° = 75° (eq. 20)
45° – 30° = 15° (eq. 21)
Therefore, let's find the values of the three primary trigonometric functions for 75° and 15° using the values of the primary trigonometric functions for 45° and 30°.
First, we know that
(eq. 22)
(eq. 23)
tan 45° = 1 (eq. 24)
(eq. 25)
(eq. 26)
(eq. 27)
We will use these numbers to find the values of the trigonometric functions for 75° and 15°. Now we can find the sine of 75° with equation 1
sin(A + B) = sin A cos B + cos A sin B, (eq. 1)
where we let A = 45° and B = 30°. So, just plug that into equation 1,
sin 75° = sin 45° cos 30° + cos 45° sin 30°. (eq. 28)
Now, substitute the values from equations 22, 23, 25 and 26 as appropriate.
(eq. 29)
Next, simplify this
(eq. 29a)
(eq. 29b)
Okay, now we need to use our calculator to check this. First use your calculator to evaluate the right hand side of equation 29b. Make sure you use parenthesis as appropriate so that you calculate the correct expression. You should find
(eq. 29c)
Now, use your calculator to find the sine of 75°. You should get
LHS = sin 75° = 0.9659258263.... (eq. 29d)
The righthand side is equal to the lefthand side. Therefore, we can say confidently that equation 1 is true for the values A = 45° and B = 30°. I will let you try as many additional values as you wish to convince yourself that equation 1 is true in general.
￼￼￼
Next, we wish to find the cosine of 75°. To do this, we need to use equation 3,
cos(A + B) = cos A cos B – sin A sin B (eq. 3)
Substitute A = 45° and B = 30° into equation 3.
cos 75° = cos 45° cos 30° – sin 45° sin 30° (eq. 30)
Substitute the values from equations 22, 23, 25 and 26.
(eq. 31)
Again, simplify this
(eq. 31a)
(eq. 31b)
Finally, use your calculator to evaluate the righthand side and left hand side separately. This gives us
(eq. 31c)
LHS = cos 75° = 0.2588190451.... (eq. 31d)
These are the same. Therefore equation 3 is true for the values A = 45° and B = 30°. Again, you can try as many additional values as you wish to convince yourself that equation 3 is true in general.
Finally, let's find the value of the tangent of 75°. We need to use equation 5 for this.
(eq. 5)
Substitute A = 45° and B = 30° into equation 5,
(eq. 32)
Then substitute the values from equations 24 and 27 into equation 32,
(eq. 33)
Finally, simplify this
(eq. 33a)
(eq. 33b)
(eq. 33c)
Next, rationalize the denominator,
(eq. 33d)
(eq. 33e)
(eq. 33f)
(eq. 33g)
Alright, all we need to do is use our calculator to compare the righthand side and the lefthand side of the equations.
(eq. 33h)
LHS = tan 75° = 3.732050808.... (eq. 33i)
As we expected, these are equal. We already demonstrated that equation 5 is true in general, as long as equations 1 and 3 are true in general. Therefore, you do not need to try any other values, but you can if you wish.
Okay, that demonstrates all the sum identities, and this is enough for one section.
The first set of advanced identities that we will be considering are called the sum and difference identities. These are called this because they take the trigonometric functions of a sum or difference of two angles, and write them in terms of the trigonometric functions of these angles.
Recall, before calculators people made tables of the trigonometric functions and then interpolated the tables for angles between the calculated points. You can imagine that these identities were very useful for checking the interpolation and calculation algorithms. By using these identities repeatedly you would be able to know precisely the trigonometric functions of many more angles than just the dozen special angles we know now.
The identities are
sin(A + B) = sin A cos B + cos A sin B (eq. 5)
sin(A – B) = sin A cos B – cos A sin B,
cos(A + B) = cos A cos B – sin A sin B, (eq. 5)
cos(A – B) = cos A cos B + sin A sin B, (eq. 5)
, (eq. 5)
. (eq. )
You do not need to memorize these identities. You do need to be able to know them on demand. Therefore, you will want to write them down on your formula sheet for the class, and keep them with your while doing homework.
With a little bit of algebra, we can derive equation 5 from equations 1 and 3, and equation 6 from equations 2 and 4 using the tangent ratio identities. Let's derive one of these, for practice. We know that the tangent of A + B is the sine of A + B divided by the cosine of A + B, so start with that,
(eq. 7)
Now replace equation 1 for the numerator in equation 7 and equation 3 for the denominator,
(eq. 8)
We wish for the righthand side of the equation to be in terms of tangents, not sines and cosines. The only thing we can do is multiply by 1, but 1 has many different forms. For example,
(eq. 9)
Multiply equation 8 by equation 9,
(eq. 10)
Separate the fractions,
(eq. 11)
Okay, we have four expressions in equation 11 that we can simplify. Let's look at these one at a time. The top left expression (we will call this the TLE, even though that isn't a real mathematical term) is
(eq. 12)
We can see that sin A divided by cos A equals tan A, and cos B divided by cos B equals 1. Therefore, the top left expression simplifies to
TLE = tan A (eq. 13)
The top right expression is
(eq. 14)
Again we can simplify this, cos A divided by cos A equals 1, sin B divided by cos B equals tan B. Therefore the top right expression is
TRE = tan B. (eq. 15)
The bottom left expression is
(eq. 16)
Clearly this simplifies to 1. Therefore the bottom left expression is
BLE = 1 (eq. 17)
Finally, the bottom right expression is
(eq. 18)
sin A divided by cos A equals tan A, and sin B divided by cos B equals tan B. Therefore, the bottom right expression is equal to
BRE = tan A tan B (eq. 19)
Finally, substitute all the pieces back into equation 11 and we get
(eq. 5)
which you might recognize as equation 5. Tuh duh!
Now that you have some experience with the basic identities and solving trigonometric problems, it is time to expand your toolkit and add a few more identities.
You may be wondering why you need to do this at all. One reason is that we need one more identity to solve our unhappy fowl problem. Another reason is that this is a precalculus class. It is not unreasonable to assume that you are taking this class to prepare to take calculus sometime in high school or college. In my mathematical career, the primary thing I used advanced trigonometric identities for was to solve calculus problems. Like it or not, you need trigonometry to do calculus.
Of course, you realize there is a finite number of special angles, therefore, a finite number of equations that can be solved by looking for the special angles that solve the equation. In the real world, there are potentially an infinite number of trigonometric equations that we can solve. Clearly, most of the equations you may runacross will have arbitrary angles as solutions. So, how do we solve these problems?
Consider the following equation,
15 sin^{2} x – 7 sin x – 2 = 0 (eq. 46)
The first thing we need to do is factor this. You can use whatever factoring technique you prefer. This can be solved with trial and error. 15 factors into 3 × 5, 2 factors into 1 × 2. Therefore, one possible thing we could try is
(5 sin x – 2)(3 sin x + 1)
If we multiply this out we get
15 sin^{2} x – sin x – 2
Clearly that is not the right factor. Let's try again,
(5 sin x – 1)(3 sin x + 2).
Multiply this out
15 sin^{2} x + 7 sin x – 2
Close. Our signs are wrong. Switch these and we get
(5 sin x + 1)(3 sin x – 2)
Multiply this out
15 sin^{2} x – 7 sin x – 2.
￼￼￼￼
Perfect! Okay, so equation 46 factors to
(5 sin x + 1)(3 sin x – 2) = 0. (eq. 47)
Set each of these factors equal to zero.
5 sin x + 1 = 0 (eq. 48)
5 sin x = 1 (eq. 49)
(eq. 50)
(eq. 51)
x = 11.53...° (eq. 52)
or
x = 0.201... rad (eq. 53)
3 sin x – 2 = 0 (eq. 54)
3 sin x = 2 (eq. 55)
(eq. 56)
(eq. 57)
x = 41.81...° (eq. 58)
or
x = 0.729... rad (eq. 59)
Alright? So that is half the solutions. Since this is a sine function, the other 2 solutions are found by taking 180° or π radians and subtracting x. (Feel free to double check this with the special angles.) This gives us two additional solutions,
x = 180° – (11.53...°) (eq. 60)
x = 191.53...° (eq. 60a)
or
x = π – (0.201...) (eq. 61)
x = 3.342... rad (eq. 61a)
and
x = 180° – (41.81...°) (eq. 62)
x = 138.18...° (eq. 62a)
or
x = π – (0.729...) (eq. 63)
x = 2.411... rad (eq. 63a)
Of course, we should double check these answers by plugging them back into equation 46. Did they all work? Great!
Let's list the solutions we have so far.
x = 11.54°, 41.81°, 191.54°, and 138.19° (eq. 64)
or
x = 0.201, 0.730, 2.412, and 3.343 (eq. 65)
where these answers have been roundedoff. So, are we good? Not quite? Why not? Oh yeah, we are looking for all the solutions between 0° and 360°, or 0 and 2π. 11.54° and 0.201 are not between 0° and 360°, and 0 and 2π, respectively. So what? We are simply on the wrong turn of the circle. To find the same angle on the correct turn of the circle, just add 360° or 2π, respectively. 11.54° + 360° is 348.46° and 0.201 + 2π is 6.082. So finally,
x = 41.81°, 191.54°, 138.19°, and 348.46° (eq. 64a)
or
x = 0.730, 2.412, 3.343, and 6.082 (eq. 65a)
Tuhduh! Alright, we have now found all the solutions for the intervals of interest. Just to double check that we have found all the solutions, let's graph these. Hopefully, by now you know the drill.

The graphs above can be created by entering the function
ƒ(x) = 15 sin^{2} x – 7 sin x – 2 (eq. 66)
into a graphing calculator set to “zoom trig.” By using the trace function, you should be able to verify the answers in equations 64a and 65a.
Alright. Are you ready to try this on your own?
Now that we can find all the solutions for simple trig problems, whether with special angles, or arbitrary angles, it is time to up the ante. What if you have a more complicated trigonometric problem? Something like a polynomial, with one of the trigonometric functions instead of just a polynomial in x? The simplest form of this type of polynomial is something like
2 cos^{2} x – 1 = 0. (eq. 20)
Most students will want to solve this problem in the following way. First, they will add 1 to both sides,
2 cos^{2} x = 1 (eq. 21)
Next, divide both sides by 2.
(eq. 22)
Take the square root of both sides,
(eq. 23)
Rationalize the denominator,
(eq. 24)
Then find the angle for which the cosine of that angle is . Remember, the cosine of the angle is the xcoordinate. Looking at the unit circles from above, we see that red circle has 2 points along it where the xcoordinate is : the point in the first quadrant, and the point in the fourth quadrant. Now, look on the circles with the angles to find the value of the angle that is the same as the points on the red circle where the xcoordinate is . Consider the circles below.

We see that these points correspond with the angles and in radians or 45° and 315° in degrees. Therefore, the solutions to equation 20 are
(eq. 25)
or
x = 45°, 315°. (eq. 26)
Let's check this. Again, get out your graphing calculator. Graph the function,
ƒ(x) = 2 cos^{2} x – 1. (eq. 27)
Your calculator can be set either to degrees or radians, but the graphing window needs to be set to “zoom trig.” You should get something like the following images.

How many zeros do these graphs show between 0° and 360°, or 0 and 2π? I count four. We only found two. Where did we go wrong?
The problem was with our use of the square root. It is true that ^{2} = ()^{2}. But when we take the square root function, we only pick up the positive root. If we need to find both roots we need to use the equation
(eq. 28)
A better way to solve equation 20 is to treat it like the polynomial it is. Recall, we find the zeros of a polynomial be factoring the polynomial, then setting each factor equal to zero. For example, if we had the polynomial
ƒ(x) = x^{2} – 4 (eq. 29)
To find the zeros of this polynomial, the first thing we do is factor this. Recall that the difference of two perfect squares factors in the following way:
ƒ(x) = (x + 2)(x – 2) (eq. 30)
To find the zeros of this, we set each factor equal to zero,and solve for x. This gives us
x + 2 = 0 (eq. 31)
x = 2 (eq. 32)
and
x – 2 = 0 (eq. 33)
x = 2 (eq. 34)
Two solutions. We remembered to find them both, since we solved the problem by factoring. Remember, in a polynomial, if we include all zeros: real, complex, and multiple, there are as many zeros as the order of the polynomial. For the simple trigonometry problems we are solving in this lesson, there are usually twice as many zeros in the interval from 0° to 360°, or from 0 to 2π as the order of the polynomial.
Let's go back to equation 20.
2 cos^{2} x – 1 = 0 (eq. 20)
Instead of solving for this directly, begin by factoring the polynomial,
(√2 cos x+1)(√2 cos x−1)=0 (eq. 35)
Now set each factor equal to zero, and solve those equations.
(eq. 36)
(eq. 37)
(eq. 38)
(eq. 39)
and
(eq. 40)
(eq. 41)
(eq. 42)
(eq. 43)
We now know that we need to find the angles for which the cosine is equal to and the angles for which the cosine is equal to . We already observed that the angles and in radians or 45° and 315° in degrees correspond with the points in the first quadrant, and the point in the fourth quadrant. That gives us the two solutions from 0° to 360°, or 0 to 2π for equation 43 above.
Now we need to find the solutions to equation 39. To do this, we again look at the unit circles, find the points where the xcoordinate is , then find the corresponding angles on the other circle. The two points with the xcoordinate are in the second quadrant and in the third quadrant. These points correspond with the angles and in radians or 135° and 225° in degrees.
Therefore, all of our solutions between 0° and 360°, or 0 and 2π for equation 20 are,
(eq. 44)
or
x = 45°, 135°, 225°, and 315° (eq. 45)
If we again graph equation 27 on our calculator, and use the trace function to locate the zeros, we find that the zeros are in fact the ones we have found. Yeah us!
In your homework, and in real life, sometime you will have a problem in which the solution is a special angle. However, most of the time in real life, the problem will not be a special angle. What should you do then? Consider the following problem,
tan x – 5 = 0 (eq. 9)
We actually don't need to go much further to solve this. Add 5 to both sides,
tan x = 5 (eq. 10)
... and that is it. Now, if we look at our unit circle, none of the angles have a ratio of ycoordinate to x coordinate that is equal to 5. So, what can we do? Well, the obvious thing is to take the arctangent of both sides.
arctan(tan x) = arctan(5) (eq. 11)
x = 78.69...°
or
x = 1.373... rad. (eq. 13)
(The units for equation 13 are unnecessary, radians are unitless. But, sometimes it is useful write radians, especially if the answer is not in terms of π.) So, are we done? As before, lets try graphing this.

The graphs above are found by graphing the function
ƒ(x) = tan x – 5 (eq. 14)
Again, you can do this in degrees or radians, but you must set your “zoom” to “trig.”
Notice that there are two xintercepts in the interval from 0° to 360°, or 0 to 2π. By using the arctangent function you only found one of these solutions. Ugh! what to do?
If you use your trace function, you should find that the second zero is near 260° or 4.5 radians. Great, can you find the actual zero?
Think about the properties of the tangent function. The tangent repeats its behavior every 180° or π radians. Since it is only increasing, you just have to catch the zero once each period. This means that the second zero in the interval 0° to 360°, or 0 to 2π, is the value of x you found with your calculator, plus 180° or π radians. 78.69° + 180° = 258.69° and 1.373 + π = 4.515. Therefore, all the possible solutions from 0° to 360°, or 0 to 2π are:
x = 78.69°, 258.69° (eq. 14)
or
x = 1.373, 4.515 rad, eq. 15
where the answers are rounded off to a reasonable number of decimal places. If we plug these values back into the equation we find
tan(78.69...°) – 5 = 0 (eq. 16)
tan(258.69...°) – 5 = 0 (eq. 17)
tan(1.373...) – 5 = 0 (eq. 18)
tan(4.514...) – 5 = 0 (eq. 19)
Okay? So, you can check to make sure that your answers are correct. Note, if I round off the answers and put that into my calculator I get
tan(78.69°) – 5 = 0.00003064 (eq. 16a
tan(258.69°) – 5 = 0.00003064 (eq. 17a)
tan(1.373) – 5 = 0.010399 (eq. 18a)
tan(4.515) – 5 = 0.00017107 (eq. 19a)
As you can see, the error associated with roundingoff answers of this type is fairly large. This is a larger error than you are used to seeing with standard algebra. Moral to the story, when you report your answers in the homework, you should roundoff to a reasonable value. When you are using that answer to make additional calculations, do NOT roundoff. Keep all the decimals, use the “store” button on your calculator to keep any digits that your calculator may have calculated but not displayed, etc. Do NOT roundoff.
Back to solving problems in general. If your function is a tangent function, you add 180° or π in order to find the second value. The sine and cosine function goes up and down. I don't think adding 180° or π will actually pick up the second solution. So what should you do? Think about it. Use the graphing program on your calculator. Look at the behavior of the numbers as they go around the unit circle. You will need to figure this one out.
We will eventually get back to the problem of the projectile. In the meantime, we should look at how to go about solving a simple trigonometry problem. Consider the following problem.
ƒ(x) = 2 sin x – 1. (eq. 3)
This is a basic trigonometric problem. We have replaced the term θ with an x, just to remind you that this is a variable, and also to emphasize that this is a function. It would have been equally valid to write
ƒ(θ) = 2 sin θ – 1. (eq. 3a)
Now, if I want to find the zeros of equation 3, I simply set ƒ(x) equal to zero and solve for x.
0 = 2 sin x – 1. (eq. 4)
That looks a bit odd. Let's reorder this so that it looks more like a standard equation.
2 sin x – 1 = 0 (eq. 4a)
So, to solve this, we start out in the normal way. We need to isolate x. Begin by adding 1 to both sides,
2 sin x – 1 + 1 = 0 + 1, (eq 5)
2 sin x = 1. (eq. 5a)
Next divide both sides by 2.
(eq. 6)
(eq. 6a)
You may recognize this as one of the special angles. So, we can look at the unit circles (remember these from Unit 08?) and find the angle for which the sine of that angle is

Remember, the sine of the angle is the ycoordinate. The blue circle has 2 points along it where the y coordinate is : the point in the first quadrant, and the point in the second quadrant. Now, look on the circles with the angles to find the value of the angle that is the same as the points on the blue circle on the following page where the ycoordinate is
You can see that these points correspond with the angles and in radians or 30° and 150° in degrees. You should be able to find both the point in degrees and radians.

Therefore, the solutions to equation 4a are
(eq. 7)
or
x = 30°, 150°. (eq. 8)
We should check this. Do you recall from earlier in the year we learned that the zeros of a function are the xintercepts? Yes? Great! So, get out your graphing calculator, and graph the function
ƒ(x) = 2 sin x – 1. (eq. 3)
Your calculator can be either in degrees or radians, but you will need to select the “zoom trig” option. You should get something like the following images.

This image shows the graph in degrees and radians. If you use your trace function, you will find that the two zeros between 0° to 360° are 30° and 150° and these same zeros between 0 to 2π are at and . (Or close enough, your trace function might not land on those numbers exactly.) So, the graph of the function validates the solution.
Also, notice that these graphs show two more zeros than we found. Why? Try extending the domain of the graphing window. If you graph this function from 720° to 720° (4π to 4π), you will find eight xintercepts, if you graph this from 1080° to 1080° (6π to 6π), you will find twelve xintercepts, etc. If you graph this from ∞ to ∞ you will find infinitely many xintercepts. Why?
Notice that there are two xintercepts per period or cycle. The graph from 360° to 360° (2π to 2π) is two periods, or two cycles. The trigonometric functions are, by their very nature, periodic. So, you will always pick up more than one solution, and if you want to find all the solutions, there will simply be infinitely many. So what should you do?
In the homework, you will be told the range you should use to report your answers. Most of the time it will be all solutions from 0 to 2π, or 0° to 360°. So, just keep going around the circle until you get all the solutions. If the range is larger or smaller than that, adjust the number you find accordingly.
In the previous lesson, we mentioned the complicated problem of determining the launch angle of a torpedo. We will not actually solve this problem. Instead, we will investigate a similar, but more easily solvable problem of the launch angle of a projectile that travels through the air instead of the water: think cannon, catapult, or a popular video game about birds. This is a problem that past military leaders have needed to solve.
Cannons, catapults, and birds launched at pigs should be massive enough, and shaped in such a way, that that the effects of air resistance on these objects is minimal. For this reason, these projectiles behave almost like an ideal projectile. The equations that describe the motion of a projectile are
x = v cos θ t (eq. 1)
(eq. 2)
Where x is the horizontal distance, y is the vertical distance, v is the initial speed, θ is the launch angle, t is the time, and g is the acceleration of gravity, which on the Earth is a constant equal to 9.8 m/s or 32 ft/s depending on the units. Using these equations you can solve for all sorts of things. If you know the distance you want the cannon to travel and the launch angle, you can solve for the initial speed. If you know the initial speed and the launch angle, you can solve for the position after a given amount of time. Etc.
If you know the launch angle, then the terms cos θ and sin θ are constants, and this is really not a trigonometry problem at all. Of course, in a real battle, you probably can't do much about changing the velocity of your cannon, catapult, or bird. You also can't really change the time that it takes to get from point A to point B. What you can change is the launch angle. In a real scenario, we wish to figure out what the angle must be to hit a target at a known location. This is a trigonometric problem, and this is the type of problem that we focus on in this lesson.
The next thing we wish to do with these identities is prove other identities. We already went through the process of proving a fundamental identity from the definition. However, other than the fundamental identities, it is best to not use the definitions to prove identities. It is best to use the fundamental identities to prove other identities. We did one example of this, in proving the cotangent ratio identity.
I also asked you to do this to prove the third Pythagorean identity. Were you able to do so? May as well do it here. Recall, the third identity was
tan^{2} θ + 1 = sec^{2} θ (eq. 30)
and I recommended using the ratio identity to write these in terms of sines and cosines to start.

Tuhduh!
Writing proofs is similar to simplifying expressions. In a way it may be easier, since you know what your final answer should be before you start.
Want to try another one?
Prove
sin θ  csc θ = cot θ cos θ (eq. 65)
We will start this proof by converting to sines. Then we will find a common denominator. At that point, we will see where we are.

Okay, at this point we have a sin^{2} θ term. I mentioned that whenever we see a squared term to use the Pythagorean theorem. Well...
Clearly the form we want is
sin^{2} θ + cos^{2} θ = 1 (eq. 26)
Solve this for sin^{2} θ,
sin^{2} θ = 1  cos^{2} θ (eq. 66)
Substitute that into our proof.

Good enough? Think you can do this on your own?
Hint: It is a good idea to have a list of the identities in front of you while doing your homework. If you get stuck, you can look at your list and see if anything looks the same. Good luck!
Ultimately we want to be able to use identities to write proofs and solve problems. To do this, we need to be able to use the fundamental identities to simplify expressions.
When we simplify an expression, the goal is to reduce the number of terms, and rewrite all terms as functions of θ, as opposed to functions of 2θ , or θ + 90°, for example. It is often possible to simplify an expression in more ways than one.
Consider the following expression:
We wish to simplify this. First, since these are functions of θ  90°, we want to find an identity that will change these to functions of θ. The cofunction identities will change a function of 90°  θ to a function of θ. However, we do not have 90°  θ, we have θ  90°. On the other hand, we know that
 b = (b  ) (eq. 50)
Therefore
θ  90° = (90°  θ) (eq. 51)
So make that substitution,
(eq. 52)
We still do not have functions of 90°  θ, but if we let x = 90°  θ, we can use the symmetry identities. Start by making that substitution,
(eq. 53)
Now we can use the symmetry identities:
sin(θ) = sin θ (eq. 44)
cos(θ) = cos θ (eq. 45)
therefore
(eq. 54)
Next, replace x with 90°  θ,
(eq. 55)
At this point, these are in the correct form to use the cofunction equations:
sin θ = cos(90°  θ) (eq. 24)
cos θ = sin(90°  θ) (eq. 25)
If we make those substitutions we find
(eq. 56)
Finally, we can use the cotangent ratio identity,
(eq. 19)
to simplify this:
(eq. 57)
This is a single term which is a function of θ. We cannot simplify this any more. Putting this together we get
(eq. 58)
To simplify this we used symmetry identities, cofunction identities, and a ratio identity. We could have started by taking the ratio identity, and worked in that direction instead. Wish to try another one? Consider the following expression
sin θ tan θ + cos θ
In the previous problem, it was clear we would need to use the cofunction identity eventually, since the terms were not functions of θ. Instead, we had an additional 90°. As we tried to write these in the correct form, it was clear we needed to use the symmetry identities. As we took each step, the next one became clear.
In this problem, it is less obvious where to start. A good rule of thumb is, when in doubt, convert things to sines and cosines. So, start by using the tangent ratio identity,
(eq. 13)
If we make this substitution we get
(eq. 59)
If we multiply the sine terms we will get
(eq. 60)
Next, find a common denominator,
(eq. 61)
(eq. 61a)
(eq. 61b)
At this point, you should be able to see which identity to use next. Anytime you have a squared term, you should look for a way to use the Pythagorean theorem. In this case, we already have sin^{2} θ + cos^{2} θ, so we can just go ahead and substitute
sin^{2} θ+ cos^{2} θ = 1 (eq. 26)
into the expression we are trying to simplify,
(eq. 62)
Sometimes, you may just have one squared term. If this is the case, you can solve the applicable version of the Pythagorean theorem for the term you have, and substitute that.
￼￼Okay, back to our problem. We can use the reciprocal identity
(eq. 15)
If we make this substitution we get
(eq. 63)
Putting this together we find
sin θ tan θ + cos θ = sec θ (eq. 64)
Cool?
Most of the problems in your homework will not simplify to a single term. You will have to use your best judgment to decide when you are done.
There are a couple remaining fundamental identities: the periodic identities and the symmetry identities. These are both properties of the functions as θ moves around the unit circle. We discovered these in the previous unit. Therefore, we will not be deriving these here. We will however list them.
The periodic identities come from the periodic nature of the trig functions. Recall that the sine, cosine, cosecant and secant functions have periods of 360° = 2 π, and the tangent and cotangent functions have periods of 180° = π.
Therefore, the values of these functions are repeated every period. This means the periodic identities are:
sin θ = sin(θ + 360°) (eq. 36)
sin θ = sin(θ + 2 π) (eq. 36a)
cos θ = cos(θ + 360°) (eq. 37)
cos θ = cos(θ + 2 π) (eq. 37a)
tan θ = tan(θ + 180°) (eq. 38)
tan θ = tan(θ + π) (eq. 38a)
csc θ = csc(θ + 360°) (eq. 39)
csc θ = csc(θ + 2 π) (eq. 39a)
sec θ = sec(θ + 360°) (eq. 40)
sec θ = sec(θ + 2 π) (eq. 40a)
cot θ = cot(θ + 180°) (eq. 41)
and
cot θ = cot(θ + π) (eq. 41a)
The symmetry identities are are based on the symmetry of the function. Recall the sine, the tangent, the cosecant and the cotangent functions have odd symmetry and the cosine and secant functions have even symmetry. We noted that if a function has odd symmetry then if you reflect the function across the origin, you return the original function. Mathematically, this means that
ƒ(x) = ƒ(x) (eq. 42)
If a function has even symmetry, if you reflect the function across the yaxis you return the original function. Mathematically, we write this as
ƒ(x) = ƒ(x). (eq. 43)
Therefore, the symmetry identities are:
sin(θ) = sin θ (eq. 44)
cos(θ) = cos θ (eq. 45)
tan(θ) = tan θ (eq. 46)
csc(θ) = csc θ (eq. 47)
sec(θ) = sec θ (eq. 48)
and
cot(θ) = cot θ (eq. 49)
So, these are the fundamental identities. What do can we do with them?
The final set of identities we can prove/derive using only the definitions are the Pythagorean identities. I will prove the first one, then use that to derive the second one. From there you can decide if you wish to derive the third identity, or prove it.
The first, and principle, Pythagorean identity is
sin^{2} θ + cos^{2} θ =1 (eq. 26)
I should mention notation at this point. sin^{2} θ is the standard way to write (sin θ)^{2}. It is written like this because it is cleaner to write sin^{2} θ than to write (sin θ)^{2}. If we were to write sin θ^{2}, this means that the term θ is squared, not the expression sin θ. In equation 30, it is the entire expression sin θ that is squared. Likewise, the expression cos θ is squared. If you see anything else, such as tan^{2} θ, this means that tan θ is squared.
Okay, back to the Pythagorean identity. This is called the Pythagorean identity because it is a form of the Pythagorean theorem. We will use the Pythagorean theorem to prove it.

You may not be familiar with the Pythagorean theorem written like (pop)^{2} + (adj)^{2} = (hyp)^{2}, but that is the what we really mean when we say ^{2} + b^{2} = c^{2}. is one leg, b is the other and c is the hypotenuse. opp is merely one leg, adj is simply another, and hyp is obviously the hypotenuse.
Okay, now that we have one form of the Pythagorean identity, I will derive the second form. Start with equation 26.
sin^{2} θ+cos^{2} θ = 1 (eq. 26)
Divide this equation by sin^{2} θ.
(eq. 27)
(eq. 27a)
(eq. 27b)
Next, use the cotangent ratio identity and the cosecant reciprocal identity to simplify this
1 + cot^{2} θ = csc^{2} θ (eq. 28)
Often you see this written as
csc^{2} θ  cot^{2} θ = 1 (eq. 29)
It doesn't matter how you want to write this, but I find it easier to remember equation 28.
The final Pythagorean identity is
tan^{2} θ + 1 = sec^{} θ (eq. 30)
You also see this written as
sec^{2} θ  tan^{2} θ = 1 (eq. 31)
While these are not the same equation, they represent the same identity.
Okay, now it is your task to either prove this, or derive it. If you wish to derive it, start with equation 26 and divide by cos^{2} θ. If you wish to prove this, use the reciprocal and ratio identities to write it in terms of sines and cosines.
Answers to section 1.4
tan θ = cot(90°  θ) (eq. 32)
(eq. 32a)
csc θ = csc(90°  θ) (eq. 33)
(eq. 33a)
sec θ = sec(90°  θ) (eq. 34)
(eq. 34a)
cot θ = tan(90°  θ) (eq. 35)
(eq. 35a)
If we are clever, we can also derive the cofunction identities from the definitions. Recall that for any triangle, the sum of the internal angles is equal to 180°. In a right triangle, one of the angles is equal to 90°. Therefore, the sum of the other angles is equal to 90°. You may recall that if two angles sum to 90°, these angles are complementary. In a right triangle, the nonright angles are complementary.
Now, consider our triangle from section 1.1, next page. Since ∠θ and ∠φ are the nonright angles in this triangle,
θ + φ = 90° (eq. 20)

Solving this for φ we get
φ = 90°  θ (eq. 21)
Recall, the trigonometric functions of θ and φ, as read off the triangle, are
(eq. 1a)
(eq. 2a)
(eq. 7a)
and
(eq. 8a)
From this we can see that equation 1a is equal to equation 8a, and that equation 2a is equal to equation 7a. If we set these equal we get
sin θ = cos φ, (eq. 22)
and
cos θ = sin φ (eq. 23)
Finally, substitute equation 20 into these and we get
sin θ = cos(90°  θ) (eq. 24)
and
cos θ = sin(90°  θ) (eq. 25)
These are the cofunction identities of the sine and cosine.
Using the ratio identities and the reciprocal identities, can you derive the cofunction identities of the remaining four trigonometric functions? Answers will be at the end of the next section.
BTW, a derivation differs from a proof primarily in the starting point. If we start with the final answer, we have a proof. If we end with the final answer we have a derivation. Another difference is that a derivation allows you to manipulate the equation, while a proof requires you to only manipulate each expression. However, you should have no trouble turning a derivation into a proof.
One more thing. Since you are also expected to be able to use radians, you should recognize that . Therefore, the cofunction identities can also be written as
(eq. 24a)
and
(eq. 25a)
You need to be able to use either equation.
The next group of identities we can determine from the definitions is the reciprocal identities. You may have noticed the following relations: Equation 4 is the reciprocal of equation 1. Equation 5 is the reciprocal of equation 2. Equation 6 is the reciprocal of equation 3. These relations are true for equations 712.
Writing these in mathematical notation we find
(eq. 13)
(eq. 14)
(eq. 15)
(eq. 16)
(eq. 17)
and
(eq. 18)
These are the reciprocal identities.
￼I mentioned we can prove these from the definitions. So, we should do that. Start with the cosecant. Because it is more logical, I will work the RHS.

I assume you can write proofs of the other reciprocal identities?
Technically, we have not proven all the reciprocal identities, but the proof for each is just a variation of the identity we proved. Therefore, go ahead and use these identities as proven identities.
One thing we can do with these identities is the prove the cotangent ratio identity. We could have used the definitions to prove this, but for practice, let's use the cotangent reciprocal identity. The cotangent ratio identity is
(eq. 19)
We can work this anyway we wish. But since it is better form, let's start from the LHS. If we get stuck, we can always work the other side.

... and that is how you write a proof without using the definitions!
Perhaps the most fundamental identity is the tangent ratio identity. This is a very important identity. You will use it all the time!
The tangent ratio identity is
(eq. 13)
This can be easily proven from the definition. So, this is a good time to introduce you to trigonometric proofs.
In a trigonometric proof, you write the identity you wish to prove, then construct 2 columns beneath it. In the left column, you manipulate the lefthand side (LHS) of the equation, until it equals the right hand side of the equation. In the right column, you manipulate the righthand side (RHS) of the equation, until it equals the lefthand side. You can work either the LHS, the RHS, or both. It is best form to work only the LHS, but not required.
In trigonometric proofs, it is common to not justify each step. However, it is good form to do so. To justify each step, you will need a third column. In this third column you will list the identity that allowed you to go from one line to the next. If you work both sides, you may need four columns. To justify a step, you can use algebraic manipulations, definitions, or previously proven identities.
That is the background. Now, let's prove the tangent quotient identity as an example of a trigonometric proof. Let's do this three times, working the LHS, the RHS and both sides.
It really is easiest to work both sides at once, so we will do that first. You want to write the identity you wish to prove above the columns, with the equal sign centered over the middle line. The "justification" columns are on either side of the center columns.

You are finished when the LHS is equal to the RHS. QED is short for quod erat demonstrandum, which is Latin for "that which was to be demonstrated." You end your proof with this because it makes you feel awesome and look cool!
The both column method has the advantage that if you get stuck, you can work the other side. It is best to only work one side, and the LHS is preferred to the RHS. With the above proof, it is a bit easier to see how the RHS transforms to the LHS, so I will show you that proof next.

We can work this from the LHS as well, which is the preferred technique. It will require a bit more ingenuity.

Notice, in all three proofs we took the same steps. The only thing that varied was the order. The reason it was trickier to work the LHS is that the idea of multiplying by that particular value of one, or converting multiplication of a fraction into division by a fraction is not something you might normally think to do. On the other hand, if you had already written the proof by working both sides, you would know where you needed to go. Therefore, you would be able to figure out less obvious steps you need to take to get there.
Now have a proven identity. We can use it to prove other identities.
In the previous lesson, you proved that you were able to determine the trigonometric functions using a a variety of techniques. However, let's take a moment to review the trigonometric functions as defined by the ratios of the sides of a right triangle.
Consider ΔABC, shown here. This is labeled in the standard way. The vertices are labeled with the capital letters A, B and C, such that ∠C is the right angle. The sides are labeled so that leg is opposite ∠A, leg b is opposite ∠B, and the hypotenuse is c. It is fairly common to use a Greek letter to name the angle, instead of the vertex. So, in the image, ∠A has been named ∠θ (theta, pronounced 'thayta') and ∠B has been named ∠φ (phi, pronounced 'fie'). The arcs drawn in the angles are part of the label.

The ratios of the sides of this triangle are dependent on only one of the nonright angles. Consider ∠θ, the measure of ∠θ is θ. The length of the hypotenuse is c, the length of the leg opposite the ∠θ is , and the length of the leg adjacent to ∠θ is b. Often we will not have the triangle fully labeled, but just refer to these sides as hypotenuse, opposite, and adjacent.
With these three sides, we define the following functions
(eq.1)
(eq.2)
(eq.3)
(eq.4)
(eq.5)
and
(eq.6)
What if we wanted to consider ∠φ, measure φ, instead? In this case, then the hypotenuse still has length c, but now the opposite leg has length b, and the adjacent leg has length , so the trigonometric functions are
(eq.7)
(eq.8)
(eq.9)
(eq.10)
(eq.11)
and
(eq.12)
Do you notice any relationships here?
These relationships are called identities. An identity is an equation that expresses a relationship that is always true, regardless of the value of the angle. We will learn about the fundamental identities in this lesson.
