If $\displaystyle cos a/cos b + sin a/sin b=-1$. Find the value of $\displaystyle cos^3 b/cos a + sin^3 b/sin a$.
Thanks.
Stage 1. What does the given information $\displaystyle \frac{\cos a}{\cos b} + \frac{\sin a}{\sin b}=-1$ tell you? Multiply out the fractions and it becomes $\displaystyle \cos a\sin b + \sin a\cos b = -\cos b\sin b$. Using standard trig formulas for sines and cosines , you can write this as $\displaystyle \sin(a+b) = -\frac12\sin(2b)$.
Stage 2. Can you think of any examples where that condition is satisfied? One way to get $\displaystyle \sin(a+b) = -\frac12\sin(2b)$ is if $\displaystyle \sin(2b) = 1$ and $\displaystyle \sin(a+b) = -\frac12$. That will occur for example if $\displaystyle b = \pi/4$ and $\displaystyle a = -5\pi/12$. Then $\displaystyle \cos b = \sin b = 1/\sqrt2$, and (see here) $\displaystyle \cos a = \frac{\sqrt2}4(-1+\sqrt3),\ \,\sin a = \frac{\sqrt2}4(-1-\sqrt3)$. For that example, $\displaystyle \dfrac{\cos^3 b}{\cos a} + \dfrac{\sin^3 b}{\sin a} = \dfrac1{2\sqrt2}\left(\dfrac4{\sqrt2(-1-\sqrt3)} + \dfrac4{\sqrt2(-1+\sqrt3)}\right)$, which simplifies (details omitted – do the calculation to check that it works) to 1.
Stage 3. Make a wild guess. Could it always be true that $\displaystyle \frac{\cos^3 b}{\cos a} + \frac{\sin^3 b}{\sin a} = 1$ (given that $\displaystyle \frac{\cos a}{\cos b} + \frac{\sin a}{\sin b}=-1$)?
Stage 4. See if you can prove that the guess in Stage 3 is correct. If it is, then multiplying out the fractions means that we want to show that $\displaystyle \sin a\cos^3b + \cos a\sin^3b = \cos a\sin a$. Let's see:
$\displaystyle \begin{aligned}\sin a\cos^3b + \cos a\sin^3b &= \sin a\cos b(1-\sin^2b) + \cos a\sin b(1-\cos^2b)\\ &= \sin a \cos b + \cos a\sin b - \sin b\cos b(\sin a\sin b + \cos a\cos b)\\ &= \sin(a+b) - \sin b\cos b\cos(a-b)\\ &= -\cos b\sin b + \sin(a+b)\cos(a-b)\quad\text{\footnotesize (using the given information)}\\ &= \ldots\quad\text{\footnotesize (can you complete the proof from there?)} \end{aligned}$