Prove $\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$ Im trying to prove the following equation:
$$\sin^6x = \frac1{32}(10 − 15\cos2x + 6\cos4x − \cos6x)$$
I have tried to do it several times now but get stuck at different places. I don't know if it's a calculation error or the question is defective. It would be really helpful if someone could point out the method.
we are to use $$z - \frac1{z} = 2i\sin x$$
$$z^n + (\frac1{z})^n = 2\cos nx$$
 A: Recall that
$$
\sin t=\frac{e^{it}-e^{-it}}{2i}
$$
so (recalling that $(2i)^6=-64$),
$$
-64\sin^6x=
e^{6ix}-6e^{4ix}+15e^{2ix}-20+15e^{-2ix}-6e^{-4ix}+e^{-6ix}
$$
Since $2\cos t=e^{it}+e^{-it}$ you should be able to finish up.
A: The problem is simple but you may have made a calculation mistake. Let's start with the identity $$\sin^{2}x = \frac{1 - \cos 2x}{2}$$ and then cube it to get $$\sin^{6}x = \frac{1}{8}(1 - 3\cos 2x + 3\cos^{2}2x - \cos^{3}2x)$$ and then we need to note that $$\cos^{2}2x = \frac{1 + \cos 4x}{2},\,\cos 6x = 4\cos^{3}2x - 3\cos 2x$$ and using these we get $$\sin^{6}x = \frac{1}{8}\left(1 - 3\cos 2x + \frac{3 + 3\cos 4x}{2} - \frac{\cos 6x + 3\cos 2x}{4}\right)$$ and on simplifying we get $$\sin^{6}x = \frac{10 - 15\cos 2x + 6\cos 4x - \cos 6x}{32} $$
A: Wolfram Alpha corroborates the identity so I think the question is not defective.  You probably want to apply the identities
\begin{align*}
    \sin^2(\theta) &= \frac{1}{2}\left(1-\cos 2\theta\right) \\
    \cos^2(\theta) &= \frac{1}{2}\left(1+\cos 2\theta\right)
\end{align*}
several times.
A: You can write $e^{inx}$ as either $\cos nx+i\sin nx$ or as $(\cos x + i\sin x)^n$. So that gives $\cos 6x=c^6-15c^4s^2+15c^2s^5-s^6$, writing $c=\cos x,s=\sin x$. Substituting $c^2=1-s^2$ that becomes $32s^6-48s^4+18s^2-1$. Similarly we get $\cos 4x=8s^4-8s^2+1,\cos2x=1-2s^2$. Substituting in the rhs of the given equation gives the result.
A: Although it's been solved in one of the previous answers, you could use the Binomial Theorem to expand $cos(x)+isen(x)$, knowing that $(cos(x)+isen(x))^n = cos^n(x)+isen^n(x)$ (DeMoivre's formula)
A: Raising to the $6^{th}$ power we get
$$
\eqalign{
64i^6\sin^6 x &= \bigl(z-{1\over z}\bigr)^6 \cr
&= z^6 - 6z^5{1\over z} + 15z^4{1\over z^2} -20z^3{1\over z^3} + 15z^2{1\over z^4} - 6z{1\over z^5} + {1\over z^6} \cr
&= z^6 - 6z^4 + 15z^2 -20 + 15{1\over z^2} - 6{1\over z^4} + {1\over z^6} \cr
&= z^6 + {1\over z^6} - 6\bigl(z^4+{1\over z^4}\bigr) + 15\bigl(z^2+{1\over z^2}\bigr) -20   \cr
&= 2\cos6x -6\cdot 2\cos4x +15\cdot 2\cos2x -20 \ \ \ so \ \ \ that \cr
-32\sin^6x &= \cos6x -6\cos4x +15\cos2x -10  \cr
}
$$
