Question

How do you verify `8sin²xcos²x = 1-cos(4x)`?

Answer 1

See below:

Explanation:

We know

`sin(2x) = 2 sin x cos x`

and

`cos(2x) = cos^2 x - sin^2 x = 1-2sin^2 x`

So

`8 sin^2 x cos^2 x = 2(4sin^2 x cos^2 x) = 2(2sin x cos x)^2`
`qquad = 2sin^2(2x) = 1-cos(2 times 2x) = 1-cos(4x)`

Answer 2

Please see below.

Explanation:

We have:

`8sin^2cos^2x = 1 - cos(2x + 2x)`

`8sin^2xcos^2x = 1- (cos(2x)cos(2x) - sin(2x)sin(2x))`

`8sin^2xcos^2x = 1 - (cos^2(2x) - sin^2(2x))`

`8sin^2xcos^2x= 1 - cos^2(2x) + sin^2(2x)`

`8sin^2xcos^2x = sin^2(2x)+ sin^2(2x)`

`8sin^2xcos^2x= 2sin^2(2x)`

Now recall that `sin(2x) = 2sinxcosx`, therefore `sin^2(2x) = 4sin^2xcos^2x`.

`8sin^2xcos^2x = 2(4sin^2xcos^2x)`

`8sin^2xcos^2x= 8sin^2xcos^2x`

`LHS = RHS`

As required.

Hopefully this helps!