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If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:

$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$

But could not proceed further please help.

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    $\begingroup$ It makes no sense usually to offer at this level an equation with one side that can be simplified, is it really $48/128$ on the right hand side? (Or maybe $41/128$?) Which is the source of the problem? $\endgroup$ – dan_fulea Sep 24 '18 at 10:05
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Hint:

Use the identity: $$(a-b)^2+(a+b)^2=2(a^2+b^2)$$

So:

$$\frac{2}{2}\left(\sin^8(x)+\cos^8(x)\right)=\frac{1}{2}\left((\sin^4(x)-\cos^4(x))^2+(\sin^4(x)+\cos^4(x))^2\right)$$

You can repeat this for all the terms with the form $a^2+b^2$, for the other terms simplify using trigonometric identities.

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If you define $s=\sin^2(x)$ your equation becomes $$s^4+(1-s)^4=\frac {48}{128}\\ 2s^4-4s^3+6s^2-4s+1=\frac {48}{128}\\s^4-2s^3+3s^2-2s+\frac 5{16}=0$$ for which Alpha finds the ugly real solutions $$s=\frac 12\left(1\pm\sqrt{\sqrt {11}-3}\right)$$ and the complex solutions $$s=\frac 12\left(1\pm i\sqrt{\sqrt {11}+3}\right)$$ but that doesn't seem very enlightening to me.

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Assuming $x\in \Bbb R.$ For brevity let $c=\cos x$ and $s=\sin x.$ Let $p=c^2s^2.$ We have $$ c^8+s^8=\frac {3}{8}\iff$$ $$ 1=(c^2+s^2)^4=(c^8+x^8)+c^2s^2(4c^4+6c^2s^2+4s^4)=$$ $$=\frac {3}{8}+c^2s^2(4(c^2+s^2)^2-2c^2s^2)=$$ $$=\frac {3}{8}+c^2s^2(4-2c^2s^2)\iff$$ $$\iff(0\leq p\leq 1\land \frac {5}{16}=2p-p^2)$$ $$\iff p=1- \sqrt {11}\;/4\iff$$ $$\iff |\sin 2x|=\sqrt {4p}=\sqrt {4-\sqrt {11}}.$$

Note: The 2nd, 3rd, and 4th displayed lines are a single sentence that is true iff $c^8+x^8=\frac {3}{8}.$

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Let $c=cosx$ and $s=sinx$ :

$s^8+c^8= \frac{48}{2^7}*\frac{2}{2}$

$s^8+c^8= \frac{96}{2^8} $

$(\frac{G}{2^4})^2+(\frac{H}{2^4})^2=\frac{96}{2^8}\implies G^2+H^2=96 $

$\implies s=\pm\frac{\sqrt[4]{G}}{2} \wedge c=\pm\frac{\sqrt[4]{96-G^2}}{2}$


We gonna use: $s^2+c^2=1\rightarrow c^2=1-s^2$:

$\frac{\sqrt{96-G^2}}{4}=1-\frac{\sqrt{G}}{4}$

$\sqrt{96-G^2}=4-\sqrt{G}\rightarrow$ WolframAlpha finds the solution : $G\approx9.7587$

For $\sqrt{96-g^4}=4-g$ we get real solutions : $g_1\approx0.0406-e \vee g_2\approx\pi-0.0177;$ cause $g_1^2\neq G\implies g_2$ gonna be our result of $g^2=G$ .$(g\approx 3.12389354026129)$

So:

$s=\pm\frac{\sqrt{g}}{2}$

We consider for $sinx$ in the 1st quarter:

$\frac{\sqrt{3}}{2}<\frac{\sqrt{g}}{2}<\frac{\sqrt{\pi}}{2}$

$60^\circ<x<62.4^\circ$;cause of $arcsin(\frac{\sqrt{\pi}}{2})$

Direct:

$x=arcsin(\frac{\sqrt{g}}{2})\approx62.1^\circ$


History and altenative version (with different* solution):

All starts with:

$(c^4+is^4)(c^4-is^4)=\frac{6}{2^4}$

$(\frac{A}{2^2})^2+(\frac{B}{2^2})^2=\frac{6}{2^4}\implies A^2+B^2=6 $

$\implies c^4=\frac{A}{4} \wedge s^4=\frac{\sqrt{6-A^2}}{4}$

Altenative for: $c=\pm\sqrt[4]{\frac{A}{4}} \wedge s=\pm\sqrt[4]{\frac{B}{4}}$ with $s^2+c^2=1$ :

$\sqrt{\frac{B}{4}}=1-\sqrt{\frac{A}{4}}$

$6-A^2=[2-\sqrt{A}]^2$

$6-A^2=4-2\sqrt{A}+A$

$0=A^2+A-2\sqrt{A}-2\rightarrow$ WolframAlpha finds the solution : $A\approx1.7049$

So:

$c=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}$

We consider for $cosx$ in the 1st quarter:

$x=arccos(\frac{\sqrt[4]{A}}{\sqrt{2}})\approx36.1^\circ$

*Also: If I would stand by $s=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}\rightarrow x\approx53.9^\circ$ (1st quarter)


Other possibilities via $E+F=(9)6$

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  • $\begingroup$ Just a formatting suggestion: \sin x looks better than sin x: $\sin x$ vs $sin x$. The same for the other trigonometric functions. $\endgroup$ – awkward Sep 21 '18 at 15:03
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\begin{align} \sin^8 x+\cos^8 x&=(\sin^4 x+\cos^4)^2-2\sin^4 x\cos^4 x\\ &=\left[(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right]^2-\cfrac {\sin^4 2x} 8\\ &=(1-\cfrac {\sin^2 2x}2)^2--\cfrac {\sin^4 2x} 8 \end{align} Let $t=\sin^2 2x$, we have $$(1-\cfrac {t}2)^2-\cfrac {t^2}8=\cfrac {48}{128}\Leftrightarrow t^2-8t+5=0,$$ which implies $$\sin^2 2x=4-\sqrt{11} \hspace{1in}\text{(with the other invalid root discarded)}$$

hence $$ x=\pm \cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2$$or more precisely, $$x=\pm\cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2+k\pi,\>k\in \mathbb{Z}$$

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Errata: I not thought that:

$$ (|z|(\cos\theta+i\sin\theta))^n=|z|^n(\cos(n\theta)+i\sin(n\theta)) $$

It that case (where $i^8=+1$): $$ (\cos x+i\sin x)^8=(\cos x)^8+(\sin x)^8 +R(x) $$ $$ \cos(8x)+isin(8x)=\frac{3}{8}+R(x) $$ $$ e^{i8x}=\frac{3}{8}+R(x) $$ This is a bit more into the direction you wish but more away in my opinion.

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Let $s=\sin(x)^2$ as Ross did to get $s^4+(1-s)^4=48/128$. Now to exploit symmetry let $s=z+1/2$ to get $(z+1/2)^4+(z-1/2)^4=48/128\iff 2z^4+3z^3-1/4=0$. The rest is boring routine.

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