Question

How do you prove `sin^4 X = (cos^2 2X - 2 cos 2x + 1) / 4`?

Answer 1

Using the following:

• `a^2-2ab+b^2 = (a-b)^2`
• `cos(2x) = cos^2(x)-sin^2(x)`
• `1 = cos^2(x)+sin^2(x)`

we have:

`(cos^2(2x)-2cos(2x)+1)/4 = (cos(2x)-1)^2/4`

`=((cos^2(x)-sin^2(x))-(cos^2(x)+sin^2(x)))^2/4`

`=(-2sin^2(x))^2/4`

`=(4sin^4(x))/4`

`=sin^4(x)`