Question

What is cos(2x) if sin theta = 4/5 (0 < theta < pi/2). I got cos theta to be 5/sqrt(41) and sin2theta to be 8/sqrt(41) but I don't know how to do this one.

Answer 1

You are given that `sin(x)=4/5` [with \(0<\theta<\pi/2\)], and that implies that adjacent side is 3. (Why? The hypotenuse is 5, and opposite side is 4, so... 5^2-4^2=9=3^2)

Answer 2

So, your cosine should be 3/5.

Answer 3

And for \(\cos(2\theta )\) use the following property:

\(\cos(2\theta )=1-2\sin^2\theta\)

Answer 4

(Showing, that all you needed is that property, without even having to find the cosine of theta .... )

Answer 5

@SolomonZelman My bad. For some reason I was thinking of sine as opposite divided by adjacent... aka tangent. Anyway the part I was stuck on was the 2sin^(2)theta.

Answer 6

How to prove the property, or how to plug it in?

Answer 7

Plug it in

Answer 8

\(\cos(2\theta )=1-2(\sin \theta)^2\quad \Longrightarrow \quad 1-2(4/5)^2=1-2(16/25)=1-32/25=-7/25\)

Answer 9

the last part should be .... \(\color{blue}{-7/25}\)

Answer 10

Oh ok that easy. Thanks