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### Course: Precalculus>Unit 2

Lesson 10: Using trigonometric identities

# 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)?
• Since C can be negative (when pi/2 < theta < pi) how do we know the value in the radical at will 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.
• 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.
• 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....?
• 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
• Is there any difference between 'principal square root' and 'square root'.?
Thank You
• 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.

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

Hope this helps.
• 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?
• 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.
• Just being curious,
If cos(2theta) = cos^2 (theta) - sin^2 (theta)
then is cos (theta) = cos^2 (theta/2) - sin^2 (theta/2) ?
• 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.
• 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.
• 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.
• Will this type of question appear on the SAT or ACT?
• 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!
• Where i can find the proof of this identities.?
• Hey, I'm kind of confused as to how to simplify these expressions:
sin(3x)cos(x) − cos(3x)sin(x)

sin(π/5)sin(π/3) - cos(π/5)cos(π/3)

(1- tan(80^0) tan(20^0))/(tan(80^0) + tan(20^0))

If anyone could walk me through these questions it would be fantastic! Thanks in advance