cos2θ=2-2sin^2θ

Question

cos2θ=2-2sin^2θ

anonymous · Answer

$$\cos(2 \phi)=\cos^2(\phi)-\sin^2(\phi)$$

anonymous · Answer

Use that.

anonymous · Answer

i have to solve for θ

anonymous · Answer

Yeah, use that identity. Then the thing factors.

anonymous · Answer

can you write it out i dont get it

anonymous · Answer

use cos 2theta = 1 - 2sin^2theta

anonymous · Answer

jimmy, look a this:
$$1-2\sin^2(\theta)=2-2\sin^2(\theta) \rightarrow 1-\sin^2(\theta)=2(1-\sin^2(\theta)) \implies 1=2?$$

anonymous · Answer

Its not solvable. Its a false statement.

anonymous · Answer

so how do you solve it i dont get it?

anonymous · Answer

so how do you solve it i dont get it?

anonymous · Answer

You can't solve it. Its a false statement.

anonymous · Answer

are you sure because this a take home test from college and i want to get a good grade cuz im paying

anonymous · Answer

If you type it in wolfram it says "False" I'll link it: http://www.wolframalpha.com/input/?i=cos(2x)%3D2-2sin^2(x) It even says: "No solutions exist"

anonymous · Answer

i get what jimmy said is false but what would cos2θ=2-2sin^2θ be like what would θ be?

anonymous · Answer

It wouldn't be anything. It is not true for ANY theta.