<s>[INST] Why is this trigonometric identity true? Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$. 
Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$
 [/INST] 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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