pre-calculus: establish the identity?

Question

(sin^2 θ) (cos^2 θ) = (1/8)[1 - cos (4θ)]

I don't know how to do this. We're allowed to use any trigonometric identity up to the double angles and half angles (those were the ones we just learned).

This is what I've done so far, but I don't know whether I'm heading in the right direction:

(sin^2 θ) (cos^2 θ)
= (1 - (cos^2 θ)) (cos^2 θ)
= (cos^2 θ) - (cos^4 θ)

Now what?

I don't get how Northstar got:

(1/4)[sin² 2θ]
= (1/4)[(1 - cos 4θ)/2]

Answer 1

Prove the identity.

(sin²θ)(cos²θ) = (1/8)[1 - cos(4θ)]
______________

Left Hand Side = (sin²θ)(cos²θ) = [(sinθ)(cosθ)]²

= [(1/2)(sin 2θ)]² = (1/4)[sin² 2θ]

= (1/4)[(1 - cos 4θ)/2]

= (1/8)(1 - cos 4θ) = Right Hand Side

Answer 2

You're allowed to work with either side of it. I'd suggest you work with the right side instead of the left side. You have (1/8)(1 - cos(4q)) - use the double angle formula on it, regroup and see what you have. Obviously, what you have will be in terms of 2q. You'll probably need to do it again. Once you have the right side in terms of just q, you'll probably be able to see the path more clearly.

Answer 3

cos(2x)=cos^2 x - sin^2 x = 2cos^2 x -a million = a million - 2sin^2 x sin(2x)=2sin x cos x 2tan(2x)=2sin(2x)/cos(2x)=4sin x cos x / (cos^2 x - sin^2 x) locate a hassle-free denominator of the different area: ((cos x + sin x)^2 - (cos x - sin x)^2)/ ((cos x - sin x)(cos x + sin x))= ((cos^2 x +2cos x sin x + sin^2 x) - (cos^2 x -2cos x sin x + sin^2 x))/(cos^2 x - sin^2 x)= integrate like words: 4cos x sin x / (cos^2 x - sin^2 x)= what we discovered above