Question

establish the identity sin²θcos²θ = 1/8[1-cos4θ]

Answer 1

cos4?

Answer 2

I think a good starting point would be to use the identity\[2\sin(\theta)\cos(\theta)=\sin(2\theta)\]

Answer 3

From there, you can simplify the LHS to \[{1 \over 4} \cdot \sin^2(2\theta)\]

Answer 4

sin^2(x)=1/2(1-cos4x)
thus sin^2cos^x=1/4*1/2(1-cos4x)
=1/8(1-cos4x)

Answer 5

sr: sin^2(2x)=1/2(1-cos4x)

Answer 6

Also, note that \[\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)\]From there, you should be able to do some algebraic manipulation to get prove LHS=RHS.

Answer 7

Or you could use the identity provided by chanh_chung, although that is a slightly lesser known identity.

Answer 8

Okay thank you!

Answer 9

To make his identities slightly more legible, the identities you need are \[\sin^2(\theta)={1 \over 2}\cdot \left(1-\cos(2\theta)\right) \]\[\cos^2(\theta)={1 \over 2}\cdot \left(1+\cos(2\theta)\right) \]

Answer 10

oh, not his,