<s>[INST] Show that $\frac{\sin^3 \beta}{\sin \alpha} + \frac{\cos^3 \beta}{\cos \alpha} = 1$ with certain given $\alpha, \beta$ Let $$\frac{\sin (\alpha)}{\sin (\beta)} + \frac{\cos (\alpha)}{\cos (\beta)} = -1 \tag{$1$}$$ where $\alpha, \beta$ are not multiples of $\pi / 2$. Show that
$$\frac{\sin^3 (\beta)}{\sin (\alpha)} + \frac{\cos^3 (\beta)}{\cos (\alpha)} = 1\tag{$2$}$$
I've tried to rewrite $(1)$ and insert into $(2)$ to get
$$ - 1 -  \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = 1 \\
\iff \frac{\sin^4 \beta \cos^2 \alpha + \cos^4 \beta \sin^2 \alpha}{\sin \alpha \sin \beta \cos \alpha \cos \beta} = -2$$
But I can't simplify any further, maybe there are some trig. identities I'm missing?
 [/INST] Multiplying by a common denominator gives:
$$\sin \alpha \cos \beta + \cos \alpha \sin \beta = -\cos \beta \sin \beta$$
which is the same as
$$\sin(\alpha + \beta) = -\cos \beta \sin \beta. \tag{1}$$
Now:
$$(\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin^2 \beta + \cos^2 \beta) = -\cos \beta \sin \beta$$
$$\sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin^3 \beta + \sin \alpha \cos^3 \beta + \cos \alpha \sin \beta \cos^2 \beta = -\cos \beta \sin \beta$$
$$\cos \alpha \sin^3 \beta + \sin \alpha \cos^3 \beta = -(\cos \beta \sin \beta + \sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin \beta \cos^2 \beta)$$
We want the RHS to equal $\sin \alpha \cos \alpha$. Repeatedly using $(1)$ we get:
$$-(-\sin(\alpha + \beta) + \sin \alpha \cos \beta \sin^2 \beta + \cos \alpha \sin \beta \cos^2 \beta)$$
$$-(-\sin(\alpha + \beta) + \sin \beta \cos \beta (\sin \beta \sin \alpha + \cos \alpha \cos \beta))$$
$$-(-\sin(\alpha + \beta) + (-\sin(\alpha + \beta))(\sin \beta \sin \alpha + \cos \alpha \cos \beta))$$
$$-(-\sin(\alpha + \beta) - \sin(\alpha + \beta) \cos(\alpha - \beta))$$
$$= \sin(\alpha + \beta) + \sin(\alpha + \beta) \cos(\alpha - \beta)$$
and the rest follows from MathFail's answer.
</s>