how to prove: $ \sin^4\theta =\frac{3}{8} - \frac{1}{2}\cos2\theta + \frac{1}{8}\cos4\theta$?

Question

How can I prove this?

$ \sin^4\theta =\frac{3}{8} - \frac{1}{2}\cos2\theta + \frac{1}{8}\cos4\theta$

Answer 1

Hint:

From left to right, start from thee complex exponential definition of $\sin\theta$ and $\cos\theta$: $$\sin\theta=\tfrac1{2i}(\mathrm e^{i\theta}-\mathrm e^{-i\theta}),\qquad \cos \theta=\tfrac1{2}(\mathrm e^{i\theta}+\mathrm e^{-i\theta})$$ whence $$\sin^4\theta=\frac1{16}\bigl(\mathrm e^{i\theta}-\mathrm e^{-i\theta}\bigr)^{\mkern-2mu4},$$ which you can expand by the binomial formula.

With pure trigonometry (always from left to right), use the linearisation formulæ: \begin{align} \sin^4\theta&=\Bigl(\tfrac12(1-\cos 2\theta)\Bigl)^2=\tfrac14(1-2\cos 2\theta+\cos^22\theta)\\ &=\tfrac14\Bigl(1-2\cos 2\theta+\tfrac12(1+\cos4\theta)\Bigr)=\dotsm \end{align}

Answer 2

Use that $$\cos(2x)=1-2\sin^2(x)$$ $$\cos(4x)=1-2\sin^2(2x)$$ $$\sin(2x)=2\sin(x)\cos(x)$$

Answer 3

It is easy. Use $\cos2x=\cos^2x-\sin^2x$ twice on $\cos4x$, then also $\sin2x=2\sin x\cos x$ and then trasform all $\sin^2x$ terms to $\cos^2x$ terms (or vice versa of course) and you obtain an equation with just multiples of $1,\cos^2(x),\cos^4(x)$.