How would I solve the following double angle identity.
$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$
So far my work is
$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$
But how would I proceed.
How would I solve the following double angle identity.
$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$
So far my work is
$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$
But how would I proceed.
Notice that \begin{eqnarray} \cos(2x)&=& \cos^2 x - \sin^2 x \\ &=& 2 \cos^2 x - 1.\\ \end{eqnarray} Then \begin{equation} \cos^2 x = \dfrac{1}{2}(1+\cos(2x)). \end{equation} Hence, \begin{eqnarray} \cos^4 x &=& (\cos^2 x)^2\\ &=& \left[\dfrac{1}{2}(1 + \cos(2x))\right]^2\\ &=& \dfrac{1}{4}(1 +2 \cos(2x)+ \cos^2(2x))\\ &=& \dfrac{1}{4} +\dfrac{1}{2} \cos(2x) + \dfrac{1}{4}\dfrac{1}{2}(1+\cos(4x))\\ &=& 3/8 + 1/2 \cos(2x) +1/8 \cos(4x) \end{eqnarray}
\begin{align*} \cos^4(x) &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^4\\ &= \frac{e^{4xi} + 4e^{2xi} + 6 + 4e^{-2xi} + e^{-4xi}}{16}\\ &= \frac{3}{8} + \frac{1}{2} \frac{e^{2xi} + e^{-2xi}}{2} + \frac{1}{8} \frac{e^{4xi}+e^{-4xi}}{2}\\ &= \frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x) \end{align*}