Simplification of $\cos^4(x) + \sin^4(x)$

Question

$$\begin{align} (\sin x)^4+(\cos x)^4 &= (1-\cos2x)^2/4 + (1+\cos2x)^2/4\\ &= (1-2\cos2x+(\cos2x)^2+1+2\cos2x+(\cos2x)^2)/4\\ &= (1+(\cos2x)^2)/2\\ \end{align}$$

Is this correct?

Answer 1

A faster way: \begin{align} \sin^4x+\cos^4x&=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x\\ &=1-\frac12(2\sin x\cos x)^2=1-\frac12\sin^22x \\[1ex] &=1-\frac12\frac{1-\cos 4x}2=\frac{3+\cos 4x}4. \end{align}

Answer 2

$$\cos2y=\cos^2y-\sin^2y=1-2\sin^2y=2\cos^2y-1$$ $$\implies\sin^2y= \dfrac{1-\cos2y}2 ,\cos^2y=\dfrac{1+\cos2y}2$$

First $y=x$

Apply the same $(y=2x)$ for $\cos^22x=\dfrac{1+\cos4x}2$

Answer 3

$$ \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^4+\left(\frac{e^{ix}+e^{-ix}}{2}\right)^4 = \frac{1}{2^4}\left(\left(e^{ix}-e^{-ix}\right)^4+\left(e^{ix}+e^{-ix}\right)^4\right) $$

but

$$ (a+b)^4+(a-b)^4 = 2 (a^4+6 a^2 b^2+b^4) $$

hence

$$ \sin^4 x+\cos^4 x = \frac{1}{2^3}\left(2\cos(4x)+6\right) = \frac{1}{4}\left(\cos(4x)+3\right) $$