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Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$.

Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$

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    $\begingroup$ Very curious identity. I'd wager it's related to ellipses somehow. Where did you encounter it? Is it an assignment? $\endgroup$ – Yuriy S Nov 11 '16 at 11:15
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    $\begingroup$ @YuriyS Just apply Cauchy-Schwarz and observe that the equality holds. The conclusion easily follows. $\endgroup$ – Cave Johnson Nov 11 '16 at 11:19
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    $\begingroup$ More specifically, you can use Titu's version of the Cauchy Schwarz here, which directly gives the equality condition. $\endgroup$ – Sawarnik Nov 16 '16 at 17:40
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There is a very direct way to solve this. First, use the given equation to find, for example, $\sin^2 \alpha$:

$$x=\sin^2 \alpha$$

$$\frac{1}{a} x^2+\frac{1}{b}(1-x)^2=\frac{1}{a+b}$$

Solving this easy quadratic, we get a single root:

$$x=\frac{a}{a+b}=\sin^2 \alpha$$

It follows:

$$1-x=\frac{b}{a+b}=\cos^2 \alpha$$

Now substitute these values into the second equation and see that it holds.

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  • $\begingroup$ this question from the book ' бескин trygonometry' but your solution is better then his :) $\endgroup$ – yavuz Feb 22 '18 at 10:31
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Solving the system $$\begin{cases}X^2+Y^2=1\\\dfrac{X^4}{a}+\dfrac{Y^4}{b}=\dfrac{1}{a+b}\end{cases}$$ where obviously $X$ and $Y$ are the sinus and cosinus respectively, one has $$X=-\sqrt{\dfrac{a}{a+b}}\\Y=\pm\sqrt{\dfrac{b}{a+b}}$$

This gives directly the equalities $\dfrac{X^8}{a^3}=\dfrac{a^4}{a^3(a+b)^4}$ and $\dfrac{Y^8}{b^3}=\dfrac{b^4}{b^3(a+b)^4}$ hence the result

$$\dfrac{X^8}{a^3}+\dfrac{Y^8}{b^3}=\dfrac{1}{(a+b)^3}$$

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The general expression is:$$\frac{\sin^{4n}\theta}{a^{2n-1}} + \frac{\cos^{4n}\theta}{b^{2n-1}} = \frac{1}{(a+b)^{2n-1}}~, ~~n\in \mathbb N$$

From the given relation, it can be written \begin{align}(a+b)\left(\frac{\sin^4\theta}{a}+ \frac{\cos^4\theta}{b}\right) &= \left(\sin^2\theta + \cos^2\theta\right)^2\\ \implies\sin^4\theta + \cos^4\theta + \frac ba\sin^4\theta + \frac ab\cos^4\theta &= \sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta\\ \implies~~~~~~~~~~~ \left(\sqrt{\frac ba}\sin^2\theta - \sqrt\frac ab\cos^2\theta \right)^2&= 0\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}&= \frac{\cos^2\theta }{b}\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}= \frac{\cos^2\theta }{b} &= \frac{\sin^2\theta + \cos^2\theta}{a+b}\end{align}

From this, it can be concluded that $$\sin^2 \theta = \frac a{a+b};~~~\cos^2\theta = \frac{b}{a+b} \,.$$

It's a matter of substituting the values to prove the general and the desired expression.

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    $\begingroup$ You should provide a reference for your inital statement. Otherwise you're just begging the question. $\endgroup$ – Carl Witthoft Nov 11 '16 at 19:57
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    $\begingroup$ @CarlWitthoft, This can be proved by mathematical induction as well as substituting the values we got from the given expression; I think OP can prove it. $\endgroup$ – user142971 Nov 12 '16 at 2:32

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