Question

cos2θ=2-2sin^2θ

Answer 1

\[\cos(2 \phi)=\cos^2(\phi)-\sin^2(\phi)\]

Answer 2

Use that.

Answer 3

i have to solve for θ

Answer 4

Yeah, use that identity. Then the thing factors.

Answer 5

can you write it out i dont get it

Answer 6

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

Answer 7

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

Answer 8

Its not solvable. Its a false statement.

Answer 9

so how do you solve it i dont get it?

Answer 10

so how do you solve it i dont get it?

Answer 11

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

Answer 12

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

Answer 13

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"

Answer 14

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

Answer 15

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