Having trouble applying half angle formula in this question,
Any hints would be great
sin^4Θ=3/8 - 1/2cos(2Θ) + 1/8 cos(4Θ) is a solution for any angle Θ
use half angle formula and cos^2(2Θ)=1/2 cos(4Θ) +1/2
thanks
Hello offahengaway and chips$\displaystyle \sin^4\theta = (1-\cos^2\theta)^2$
$\displaystyle =\Big(1 - \tfrac12(1+\cos 2\theta)\Big)^2$
$\displaystyle =(\tfrac12-\tfrac12\cos 2\theta)^2$
$\displaystyle =\tfrac14-\tfrac12\cos 2\theta +\tfrac14\cos^22\theta$
$\displaystyle = \tfrac14 -\tfrac12\cos 2\theta +\tfrac14\cdot\tfrac12(\cos 4\theta+1)$
$\displaystyle = \tfrac38 - \tfrac12\cos 2\theta +\tfrac18\cos 4\theta$
Grandad