Ridiculous Trigonometric Equation

For simplicity, $\sin \theta = s$ and $\cos \theta=c$

$s^8+c^8 = \dfrac{17}{32}$

We know that $s^2+c^2=1$ , so raising to power $4$ on both sides,

$\therefore s^8+4s^6c^2 + 6s^4c^4 + 4s^2c^6 + c^8=1$

$\therefore\dfrac{17}{32} + 4s^2c^2(s^4+c^4) + 6s^4c^4 = 1$

$\therefore 4s^2c^2(1-2s^2c^2) + 6s^4c^4 =1-\dfrac{17}{32}$

$\therefore 4s^2c^2 - 2s^4c^4 = \dfrac{15}{32}$

$\therefore s^4c^4-2s^2c^2+ \dfrac{15}{64}=0$

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This has roots $s^2c^2= \dfrac{2 \pm \sqrt{ 4 - \frac{15}{16}}}{2} = \dfrac{2\pm \frac{7}{4}}{2} = \dfrac{15}{8} \text{ or } \dfrac{1}{8}$

Thus $sc = \pm\sqrt{\dfrac{15}{8}} \text{ or } \pm\sqrt{\dfrac{1}{8}}$

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We discard $\dfrac{15}{8}$ because $sc \leq \dfrac{1}{2}$ always.

proof :- $(s-c)^2 \geq 0 \implies s^2+c^2-2sc \geq 0 \implies 1\geq 2sc \implies sc\leq \frac{1}{2}$

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This gives $\sin( 2 \theta) = 2sc = \pm 2\times \dfrac{1}{2\sqrt{2}} =\pm \dfrac{1}{\sqrt{2}}$

Thus $\sin 2\theta = \sin \frac{\pi}{4}$ or $\sin 2\theta = \sin \frac{-\pi}{4}$

From this we obtain the general solution

$2\theta = n \pi \pm \frac{\pi}{4}$

And because of the $\sin ^8$ and $\cos^8$ type of terms, we notice that this could simply be extended to $\displaystyle\theta = \dfrac{n\pi}{2} \pm \dfrac{\pi}{8}$

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The solutions of this in the interval $0\leq \theta \leq \pi$ are $\dfrac{\pi}{8} , \dfrac{3\pi}{8} , \dfrac{5\pi}{8} , \dfrac{7\pi}{8}$

Their sum is $2 \pi \approx \boxed{6.2831}$

${ sin }^{ 8 }\theta +{ cos }^{ 8 }\theta ={ ({ sin }^{ 4 }\theta +{ cos }^{ 4 }\theta ) }^{ 2 }-2{ sin }^{ 4 }\theta { cos }^{ 4 }\theta$

$={ ({ ({ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta ) }^{ 2 }-2{ sin }^{ 2 }\theta { cos }^{ 2 }\theta ) }^{ 2 }-\frac { { sin }^{ 4 }2\theta }{ 8 }$

$={ (1-\frac { sin^{ 2 }2\theta }{ 2 } ) }^{ 2 }-\frac { { sin }^{ 4 }2\theta }{ 8 }$

$={ (1-(\frac { 1-cos4\theta }{ 4 } )) }^{ 2 }-\frac { 1 }{ 8 } { (\frac { 1-cos4\theta }{ 2 } ) }^{ 2 }$

$={ (\frac { 3+cos4\theta }{ 4 } ) }^{ 2 }-\frac { 1 }{ 8 } { (\frac { 1-cos4\theta }{ 2 } ) }^{ 2 }$

$=\frac { { cos }^{ 2 }4\theta }{ 32 } +\frac { 17 }{ 32 } +\frac { 7cos4\theta }{ 16 }$

Since this expression is equal to $\frac{17}{32}$ hence we have :

$\frac { 17 }{ 32 } =\frac { { cos }^{ 2 }4\theta }{ 32 } +\frac { 17 }{ 32 } +\frac { 7cos4\theta }{ 16 }$

$\Rightarrow { cos }^{ 2 }4\theta +14cos4\theta =0$

$\Rightarrow { cos }4\theta =0$

$\Rightarrow \theta =\frac { \pi }{ 8 } ,\frac { 3\pi }{ 8 } ,\frac { 5\pi }{ 8 } ,\frac { 7\pi }{ 8 }$

$\displaystyle \sin^{8} x + \cos^{8} x = (\sin^{4} x + \cos^{4} x)^{2} - 2\sin^{4} x \cos^{4}x$

Now for simplicity $sinx = s, cosx = c$

$\displaystyle s^{4} + c^{4} = 1 - 2s^{2}c^{2}$

$\displaystyle (1 - 2s^{2}c^{2})^{2} - 2s^{4}c^{4} = \frac{17}{32}$

Now

Let

$s^{2}c^{2} = t$

Plugging it in the equation we get

$\displaystyle 2t^{2} - 4t + \frac{15}{32} = 0$

Solving the quadratic we get

t = $\displaystyle \frac{1}{8}$

But

t = $s^{2}c^{2}$

$\displaystyle sc = \sqrt{\frac{1}{8}}$

$\sin 2x = \pm \frac{1}{\sqrt2}$

Here

$x = \frac{n\pi}{2} \pm \frac{\pi}{8}$

As $0 \leq x \leq \pi$

We get

$x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$

Hence

$\displaystyle \frac{\pi}{8} + \frac{3\pi}{8} + \frac{5\pi}{8} + \frac{7\pi}{8} = \boxed{2\pi}$

No need to find out what $\theta$ 's exactly are. If a $\theta$ satisfies the equation and $0 \leq \theta \leq \frac{\pi}{4}$ , then $(\frac{\pi}{2} - \theta)$ , $(\frac{\pi}{2} + \theta)$ , $(\pi- \theta)$ will also be the solutions to the equation. Apparently sum of these 4 solutions is $2\pi$ .

Now let $f(\theta)=\sin ^8 \theta + \cos ^8 \theta$ , which has a domain of $[0, \frac{\pi}{4}]$ . We will find $f(\theta)$ as a strictly monotone decreasing function, with a range of $[\frac{1}{8}, 1]$ . So there is only one solution in $[0, \frac{\pi}{4}]$ , then 4 soluntions in $[0, \pi]$ . Eventually, the answer is $2\pi$ , which is 6.28.

No matter what $f(\theta)$ is, such as $\frac{17}{32}$ , $\frac{19}{32}$ , $\frac{21}{32}$ or else, as long as $\frac{1}{8} < f(\theta) < 1$ , the final answer is always 6.28.

Here's another way: write $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ . Expanding $\cos^8x+\sin^8x$ and collecting terms produces $\cos^8x + \sin^8x=\frac{1}{64}(\cos 8x+28\cos 4x+35).$ Writing $\cos8x = 2\cos^24x-1$ , setting the above equation to $17/32$ and simplifying produces this quadratic in $\cos 4x$ : $\cos^24x+14\cos 4x=0$ or $\cos 4x(\cos 4x+14)=0.$ The only solutions to this are $\cos 4x=0$ or $4x = \pi/2,3\pi/2,5\pi/2,7\pi/2,\ldots$ or $x=\pi/8,3\pi/8,4,5\pi/8,7\pi/8$ as the only solutions in the range $0\le x\le\pi$ . The sum is $2\pi$ .

It's a nice problem!