Main content

## Precalculus

### Course: Precalculus > Unit 2

Lesson 10: Using trigonometric identities- Finding trig values using angle addition identities
- Using the tangent angle addition identity
- Find trig values using angle addition identities
- Using trig angle addition identities: finding side lengths
- Using trig angle addition identities: manipulating expressions
- Using trigonometric identities
- Trig identity reference
- Trigonometry: FAQ

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Using trig angle addition identities: manipulating expressions

Sal is given that cos(2θ)=C and he uses the cosine double-angle identity in order to find an expression for sin(θ). Created by Sal Khan.

## Want to join the conversation?

- Where are the original videos on this topic? How do we know the equation cos(alpha)cos(beta)-sin(alpha)sin(beta)?(21 votes)
- Since C can be negative (when pi/2 < theta < pi) how do we know the value in the radical at4:45will remain positive?

Edit: Nevermind, I see my error. C can never be less than -1 on the unit circle, so that radical will never be negative.(15 votes)- you made a small mistake in your edit: if C<-1, the radical becomes more positive, as the formula has 1-C, not 1+C in it, so C would have to become larger than 1 for the radical to become negative.(6 votes)

- if sin(theta)=sqrt((1-C)/2)), and since C=cos2(theta) then would it be correct to assume that sin(theta) also = + or - sqrt(1/2 -cos(theta)), or even sin^2(theta)=1/2-cos(theta), cos(theta) would also equal sin^2(theta)-1/2....?(8 votes)
- Close but not quite. √((1-cos2Θ)/2) = √(1/2 - cos2Θ/2) and cos2Θ/2 is not the same as cosΘ. You can't divide out the 2 like that since you are multiplying it by Θ then taking the cos of that product. These would be true if C = 2cosΘ except for the last one which has a mess up of signs.

Since C = cos2Θ,

sinΘ = √(1/2 - cos2Θ/2)

sin^2Θ = 1/2 - cos2Θ/2 2sin^2Θ = 1-cos2Θ -2sin^2+1 = cos2Θ

cosΘ = √(cos2Θ+sin^2Θ),

cos2Θ = C = cos^2Θ - sin^2Θ therefore,

cosΘ = √((cos^2Θ-sin^2Θ)+sin^2Θ) = √(cos^2Θ) = cosΘ

You have the right idea, however, the problems with the calculations began with dividing out the 2 in cos2Θ.

I hope this helps, even though it is 3 years later :D(5 votes)

- Is there any difference between 'principal square root' and 'square root'.?

Thank You(3 votes)- Square roots can have 2 answers: a positive value and a negative value.

For example: 3(3) = 9 and -3(-3) = 9. So sqrt(9) could be 3 or -3.

The positive root is called the principal root. It is the default whenever you are working with square roots. So, if you are given: sqrt(9) = 3, it expects you to provide the principal / positive root.

To ask you to provide the negative root, there will be a minus sign in front of the radical.

For example: - sqrt(9) is asking for the negative root, so it = -3.

Hope this helps.(11 votes)

- Hi guys, I'm clearly missing something. Sal turns C=cos^2theta-sin^2theta into sqrt1-C/2. I used a different method. I did the following: I decided to move -sin^2theta to the left side and got C+sin^2theta=cos^2theta, then moving C to the right side gives sin^2theta=cos^2theta-C. Then I rooted both sides and got sintheta=costheta-sqrtC. The two results are not equivalent. I checked. A lot. I know I must be wrong but I don't know why. Could someone please tell me?(3 votes)
- The error is when you did the square root.

sqrt[cos^2theta-C] does not equal costheta-sqrt(C)

You can't do the square root against each term.

If you could, then something like sqrt(16+9) would = sqrt(16)+sqrt(9).

sqrt(16+9) = sqrt(25) = 5

sqrt(16) + sqrt(9) = 4+3 = 7

As you can see, they results are not equal.

Hope this helps.(2 votes)

- Just being curious,

If cos(2theta) = cos^2 (theta) - sin^2 (theta)

then is cos (theta) = cos^2 (theta/2) - sin^2 (theta/2) ?(2 votes)- That is correct. Also, since there are two other useful double angle formulae for the cosine function, they can be manipulated into the half-angle identities. The first is cos(2t) = 2cos^2 (t) - 1. See if you can use the same reasoning, along with a little algebra to derive the half angle formula for cosine.(2 votes)

- Will this type of question appear on the SAT or ACT?(2 votes)
- Yes, but this is also useful to know for future courses, like calculus, because trig identities are used to simplify most trig related calculus expressions. This is also important because it helps you find the trig value of any angle that can be represented as a sum or difference of the special angles (30, 45, 60 and 90). This is VERY useful to know!(2 votes)

- Where i can find the proof of this identities.?(1 vote)
- sin² ⊖ + cos² ⊖ = 1:

https://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/pythagorean-identity/v/pythagorean-trig-identity-from-unit-circle

cos (a + b) = (cos a)(cos b) - (sin a)(sin b):

https://www.khanacademy.org/math/trigonometry/less-basic-trigonometry/angle-addition-formula-proofs/v/proof-angle-addition-cosine(3 votes)

- Do these tasks still exist? I can't find this kind of a task(2 votes)
- There are some sections where there are no exercises. This appears to be one of them. They keep adding content all the time. Maybe they will get added in the future.(1 vote)

- Wow! So cos(2*theta) is basically equivalent to cos(theta+theta)!! This is extremely useful especially since we can see that on a unit circle, this would make necessary changes in the y and x values.(2 votes)
- Yes, that's correct! The double angle formula for cosine states that:
`cos(2θ) = cos(θ + θ)`

This formula is very useful in trigonometry, as it allows us to find the cosine of an angle that is twice as large as a given angle. We can also use it to simplify certain expressions involving cosine.

To see how this formula is derived, we can use the angle addition formula for cosine, which states that:`cos(α + β) = cos(α)cos(β) - sin(α)sin(β)`

Letting α = β = θ, we get:`cos(2θ) = cos(θ + θ) = cos(θ)cos(θ) - sin(θ)sin(θ)`

Using the trigonometric identities cos²(θ) = 1 - sin²(θ) and sin²(θ) = 1 - cos²(θ), we can simplify this expression to:`cos(2θ) = 2cos²(θ) - 1 = 1 - 2sin²(θ)`

So, we have derived the double angle formula for cosine. As you mentioned, this formula is useful because it helps us understand the changes in the x and y coordinates on the unit circle when we double the angle.(1 vote)

## Video transcript

Voiceover:Cosine of two theta is equal to C and theta is between zero and pi. Write a formula for sine
of theta in terms of C and so I encourage you to pause the video and try to figure this out on your own before I work through it. So assuming you had a go, so let's see if we can
work through it together. I'm going to get my scratchpad out, I have copied and pasted
the exact question right over here, and so
let's think about it. They're telling us that
cosine of two theta is equal to C, so let
me write it this way. C is equal to cosine of two theta. Now, using my knowledge of
angle addition formulas, we know for example that cosine of alpha plus beta is equal to cosine alpha cosine beta minus sine alpha sine beta. Now why would this be useful here? Well this is the sum of
just theta plus theta, and so I can rewrite this in terms of cosines and sines and maybe I can rewrite the cosines in terms of sines and then solve for the sines. So let's try to do that. So I can rewrite cosine of
two theta that's the same as cosine of theta plus
theta, is of course the same thing as two theta, and I can use the angle addition formula for cosine. This is going to be
equal to cosine of theta, times cosine of theta,
times cosine of theta, minus sine of theta, times sine of theta, time sine of theta, which is of course equal to cosine squared
theta, that's equal to cosine squared theta,
minus and what we have right over here is sine squared theta. So let's see, we've been able to rewrite C in terms of cosine squared theta and sine squared theta, but ideally we just want to write it in terms of sine theta so we can solve for sine theta. So if we can re-express
cosine theta in terms of sine. Well we already know from
the pythagorean identity that cosine squared theta plus sine squared theta is equal
to one, or we could say that cosine squared theta is equal to, I'm just going to subtract
sine squared from both sides, is equal to one
minus sine squared theta. So let me rewrite this
as one minus sine squared theta, and then of course we have minus this yellow sine squared theta, and all of this is equal
to C or we could get that C is equal to one minus
two sine squared theta. And what's useful about this is we just have to solve for sine of theta. So let's see, I could
multiply both sides by a negative just so I can
switch the order over here. So I could write this as negative C is equal to two sine squared theta minus one and just multiply both sides by a negative and then
let's see I could add one to both sides, if I
add one to both sides, and I'll go over here, if
I add one to both sides, I get one minus C is equal
to two sine squared theta, I could divide both sides by two, and then so I get sine squared theta is equal to one minus C over two, or I could write that sine of theta is equal to the plus or minus square root of one minus C over two. So that leads to a question, Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to
pause the video again in case you haven't
already figured it out, and look at the information here, and think about whether
they give us the information of whether we should be looking at the positive or negative sine. Well they tell us that theta
is between zero and pi. So if I were to draw a unit circle here, between zero and pi radians. So, this angle right over
here is zero radians, and pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So, it could be an angle like this, it could be an angle like this, it cannot be an angle like this, and we know that the sine of an angle is the Y coordinate, and so we know that for the first of second
quadrant the Y coordinate is going to be non-negative. So we would want to take
the positive square root, so we would get sine of theta, is equal to the principal root, or we could even think of it as the positive square root of one minus C over two. So let's go back to our... Make sure we can check our answers. So sine of theta is
equal to the square root of one minus capital C,
all of that over two, and we got it right.