If $\sin^8\theta+\cos^8\theta=\frac{17}{32}$, find the value of $\theta$ using de Moivre's theorem. If
$$\sin^8\theta+\cos^8\theta=\frac{17}{32}$$
find the value of $\theta$ using  de Moivre's theorem. I tried a lot exapanding using Binomial theorem and taking real part and equating it given value. 
 A: De Moivre's Theorem: $e^{ix} = cos(x) + i*sin(x)$. 
From this famous equation follow the well-known expressions for the cosine: $ cos(x) = (e^{ix}+e^{-ix})/2$ and the sine: $sin(x) = (e^{ix} -e^{-ix})/(2i)$. 
We seek the solutions to the equation $sin^8(x) + cos^8(x) = 17/32$. Substitute the cosine and sine expressions. After some calculation we obtain: $e^{8ix} + 28*e^{4ix} + 70 + 28*e^{-4ix} + e^{-8ix} = 68$.
Now use the cosine formula to rewrite this result in terms of $cos(4x)$ and $cos^2(4x)$:
$cos(4x)*[cos(4x) + 14] = 0$ 
The second term is positive for all values of $x$. Hence we must solve $cos(4x) = 0$. The result is: $x = (1 + 2k) \pi/8$ where $k = 0, 1, 2,3 ….$ With a pocket calculator you can check numerically that this result is correct.
A: HINT
According to the formula
\begin{align*}
\sin(2\theta) = 2\sin(\theta)\cos(\theta)
\end{align*}
We can rewrite the given expresion as
\begin{align*}
\sin^{8}(\theta) + \cos^{8}(\theta) & = [\sin^{4}(\theta) + \cos^{4}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [(\sin^{2}(\theta)+\cos^{2}(\theta))^{2} - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = [1 - 2\sin^{2}(\theta)\cos^{2}(\theta)]^{2} - 2\sin^{4}(\theta)\cos^{4}(\theta)\\
& = \left[1 - \frac{\sin^{2}(2\theta)}{2}\right]^{2} - \frac{\sin^{4}(2\theta)}{8} = 1 -\sin^{2}(2\theta) + \frac{\sin^{4}(2\theta)}{8} = \frac{17}{32}
\end{align*}
Therefore, if we make the substitution $y = \sin(2\theta)$, the given problem is equivalent to
\begin{align*}
& 4y^{4} - 32y^{2} + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2}) + 15 = 0 \Longleftrightarrow 4(y^{4} - 8y^{2} + 16) - 49 = 0 \Longleftrightarrow\\
& 4(y^{2}-4)^{2} = 49 \Longleftrightarrow y^{2} - 4 = \pm\frac{7}{2} \Longleftrightarrow y^{2} = 4\pm\frac{7}{2} \Longleftrightarrow y = \pm\frac{1}{\sqrt{2}} = \pm\sin\left(\frac{\pi}{4}\right)
\end{align*}
Can you proceed from here?
EDIT
According to the Newton's binomial theorem as well as the De Moivre's formula, we have
\begin{align*}
\cos(8\theta) + i\sin(8\theta) = [\cos(\theta) + i\sin(\theta)]^{8} = \sum_{k=0}^{8}\binom{8}{k}[\cos(\theta)]^{k}[i\sin(\theta)]^{8-k}
\end{align*}
As a consequence, you can obtain the value of the expression $\sin^{8}(\theta) + \cos^{8}(\theta)$ in terms of sines and cosines and solve the 
corresponding equation.
A: I try to give a solution with the given constraint.
Using de Moivre's formula we will show
$$
\begin{aligned}
\cos^4 t +\sin^4t &=\frac 1{4}\Big[\ \cos 4t+3\ \Big]\ ,
\\
\cos^8 t +\sin^8t &=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
\end{aligned}
$$
There is no prize of beauty we can win for the proof, so let us use $c:=\cos t$, $s=\sin t$ as shortcuts for an easy typing. The strategy is to write $\cos 4t$ and $\cos 8t$ in terms of an even polynomial in $c,s$, with no mixed monomials.
Let us get explicit relations:
$$
\begin{aligned}
\cos 4t
&=\text{Real part of }(\cos 4t +i\sin 4t)\\
&=\text{Real part of }(\cos t +i\sin t)^4\quad\text{(de Moivre)}\\
&=c^4-6c^2s^2+s^4\\
&=c^4-3(\underbrace{(c^2+s^2)^2}_{=1}-c^4-s^4)+s^4\\
&=4(c^4+s^4)-3\ ,\\
&\qquad\text{showing the first formula.}\\[2mm]
2c^2s^2
&=(c^2+s^2)^2-c^4-s^4=1-c^4-s^4\ ,\\
4c^4s^4
&=(1-c^4-s^4)^2\\
&=1+c^8+s^8-2c^4-2s^4+2c^4s^4\ ,\qquad\text{ so}\\
2c^4s^4
&=1+c^8+s^8-2c^4-2s^4\ ,\qquad\text{ so}\\[2mm]
\cos 8t
&=\text{Real part of }(\cos 8t +i\sin 8t)\\
&=\text{Real part of }(\cos 4t +i\sin 4t)^2\qquad\text{(de Moivre)}\\
&=\cos^24t-\sin^24t\\
&=2\cos^24t-1\qquad\text{(or going straightforward here...)}\\
&=2(4c^4+4s^4-3)^2
\\
&=\dots
\ .
\end{aligned}
$$
Some further lines are showing the claimed formula for $\cos 8t$. (The term in $c^4s^4$ can be split as above to have only monomials in $c$, or in $s$, or constant ones.)
From the given relation we have to solve the equation in $t$ (instead of $\theta$, which is too hard to type):
$$
\frac {17}{32}
=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
$$
Possibly using de Moivre, 
we write again
$\cos 8t = 2\cos^2 4t-1$ and obtain an equation of second degree in $u:=\cos 4t\in[-1,1]$,
$$
\frac {17}{32}
=\frac 1{64}\Big[\ 2u^2+28u +34\ \Big]\ .
$$
Now $17/32$ cancels on both sides, we get $2(u^2+17u)=0$, equivalently (for $|u|\le 1$) $u=0$, i.e. $4t$ is an odd multiple of $\pi/2$, i.e. $t$ is an odd multiple of $\pi/8$.
