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?
Best Answer
Eq.$(1)$ is equivalent to
$$\sin(a+b)=-\frac{1}2\sin(2b)\tag{3}$$
Now start from the LHS of Eq.$(2)$, and we will show it equals $1$
$$\begin{align} \text{LHS}=\frac{\sin^3(b)\cos(a)+\cos^3(b)\sin(a)}{\sin(a)\cos(a)}\end{align}$$
Deal with the numerator: $$\begin{align} \text{Numerator}&=(1-\cos^2(b))\sin(b)\cos(a)+(1-\sin^2(b))\cos(b)\sin(a)\\ \\ &=\sin(a+b)-\cos^2(b)\cos(a)\sin(b)-\sin^2(b)\cos(b)\sin(a)\\ \\ &=\sin(a+b)-\frac{1}2\cos(a)\cos(b)\sin(2b)-\frac{1}2\sin(a)\sin(b)\sin(2b)~~~~~~~\text{use} ~~(3)\\ \\ &=\sin(a+b)+\sin(a+b)\cos(a-b)\\ \\ &=\sin(a+b)+\frac{1}2\sin(2a)+\frac{1}2\sin(2b)~~~~~~~\text{use} ~~(3)\\ \\ &=\frac{1}2\sin(2a)=\text{Denominator} \end{align}$$