A Trigonometry + complex number Equation the question is:
Show that:
$$
[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\= 2^{n+1} \sin n \frac{(\theta - \phi)}{2}   \cos x \frac{\theta + \phi - \pi}{2}$$
I gave it a try following the pattern of the previous questions: https://ibb.co/d77by5
 A: I find that your result in the question is valid only for even values of $n$ (as well containing typographical errors; it should be $\sin^n$ and $\cos(n...)$). My solution is as follows:
First note that
$$\cos\theta-\cos\phi=-2\sin\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right) \\
\sin\theta-\sin\phi=2\cos\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right)$$
Then
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^n\sin^n \left(\frac{\theta-\phi}{2}\right) \cdot \left\{\left[-\sin\left(\frac{\theta+\phi}{2}\right) +i\cos\left(\frac{\theta+\phi}{2}\right)\right]^n + \left[-\sin\left(\frac{\theta+\phi}{2}\right) -i\cos\left(\frac{\theta+\phi}{2}\right)\right]^n\right\}$$
Now, let $a=(\theta+\phi)/2$ and convert the terms in the curly braces to exponential form,
$$\{\ \frac{}{}\}=\left\{\left[-\left(\frac{e^{ia}-e^{-ia}}{2i}\right) +i\left(\frac{e^{ia}+e^{-ia}}{2}\right)\right]^n + \left[-\left(\frac{e^{ia}-e^{-ia}}{2i}\right) -i\left(\frac{e^{ia}+e^{-ia}}{2}\right)\right]^n\right\}\\=\frac{i^n}{2^n} \left\{ \left[e^{ia}-e^{-ia}+e^{ia}+e^{-ia} \right]^n + \left[e^{ia}-e^{-ia}-e^{ia}-e^{-ia} \right]^n\right\} \\=\frac{i^n}{2^n}  \left\{ 2^n e^{ina}+(-2)^ne^{-ina} \right\}=i^n\left\{ e^{ina}+(-1)^ne^{-ina} \right\}$$
Now,
$$i^n \left\{ e^{ina}+(-1)^ne^{-ina} \right\} = 2i^{n+1} \sin\left(n\frac{\theta+\phi}{2}\right) \ \  \text{for odd} \ n \\ i^n \left\{ e^{ina}+(-1)^ne^{-ina} \right\} = 2i^{n} \cos\left(n\frac{\theta+\phi}{2}\right) \ \  \text{for even} \ n $$
And finally,
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=(2i)^{n+1} \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \sin\left(n\frac{\theta+\phi}{2}\right) \ \  \text{for odd} \ n \\$$
and
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^{n+1} i^n \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \cos\left(n\frac{\theta+\phi}{2}\right) \ \  \text{for even} \ n \\$$
These can be combined into a single equation as follows:
$$[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\theta)] \\=2^{n+1} (-1)^n \sin^n\left(\frac{\theta-\phi}{2}\right) \cdot \cos\left(n\frac{\theta+\phi-\pi}{2}\right)$$
which is very close the answer in the original question, differing only by the factor $(-1)^n$. It's no wonder that it worked for even $n$. All the results have been verified by direct numerical simulation.
A: As suggested, let $$\cos\theta-\cos\phi=R\cos A,\sin\theta-\sin\phi=R\sin A$$
so that $$S=[(\cos\theta - \cos\phi) + i(\sin \theta - \sin\phi)]^n + [(\cos\theta - \cos\phi) - i(\sin\theta - \sin\phi)]^n=\{R(\cos A+i\sin A)\}^n+\{R(\cos A-i\sin A)\}^n$$
$$=R^n\left[(\cos A+i\sin A)^n+\dfrac1{(\cos A+i\sin A)^n}\right]$$
Using de Moivre's Identity, $$S=2R^n\cos nA$$
Now $R^2=(R\sin A)^2+(R\cos A)^2=2-2\cos(\theta-\phi)=\left(2\sin\dfrac{\theta-\phi}2\right)^2$
WLOG $R=2\sin\dfrac{\theta-\phi}2$
again, $\dfrac{R\sin A}{R\cos A}=\dfrac{\sin\theta-\sin\phi}{\cos\theta-\cos\phi}$
Using Prosthaphaeresis Formulas, $\tan A=-\cot\dfrac{\theta+\phi}2=\tan\dfrac{\theta+\phi-\pi}2$
Can you take it from here?
A: Using Prosthaphaeresis Formula,
$$S=(\cos\theta - \cos\phi)+i(\sin \theta - \sin\phi)=2\sin\dfrac{\theta-\phi}2\left[-\sin\dfrac{\theta+\phi}2+i\cos\dfrac{\theta+\phi}2\right]$$
Now $-\sin\dfrac{\theta+\phi}2+i\cos\dfrac{\theta+\phi}2=\cos\dfrac{\theta+\phi+\pi}2+i\sin \dfrac{\theta+\phi+\pi}2$  as $\cos\left(A+\dfrac\pi2\right)=-\sin A,\sin\left(A+\dfrac\pi2\right)=\cos A$
Using De Moivre's formula,
$$S^n=2^n\sin^n\dfrac{\theta-\phi}2\left[\cos\dfrac{n(\theta+\phi+\pi)}2+i\sin\dfrac{n(\theta+\phi+\pi)}2\right]$$
Similarly,
$$\left[(\cos\theta - \cos\phi)-i(\sin \theta - \sin\phi)\right]^n=2^n\sin^n\dfrac{\theta-\phi}2\left[\cos\dfrac{n(\theta+\phi+\pi)}2-i\sin\dfrac{n(\theta+\phi+\pi)}2\right]$$ 
