Prove that $(1+\cos(\theta)-i\sin(\theta))^{n} =2^{n}\cos^{n}(\frac{\theta}{2})(\cos(\frac{n\theta}{2})-i\sin((\frac{n\theta}{2}))$ I should use complex number and polar form, but I don't know how to proceed. I do not know what trigonometric identity I could  use.
Also how can I reduce $\cos^{3}x+\cos^{3}2x+\dots+\cos^{3}nx$
Any clue/suggestion would be appreciated.
 A: Hint:for the first result
Use $1+ \cos \theta =2\cos^2\frac{\theta }{2}$
$\sin \theta =2\sin^2\frac{\theta }{2}\cos \frac{\theta }{2}$
And De'moivre theorem
$$(\cos \theta -i \sin \theta )^n=\cos n\theta -\sin n\theta$$
Hint: for the second result
Use $$\cos n\theta +\sin n\theta=(\cos \theta +i \sin \theta )^n=\cos^n\theta +^nC_1\cos \theta (i \sin \theta )+^nC_2\cos^2\theta( i\sin \theta )^2+...+(i\sin^n\theta )$$
Separate real  and imaginary parts.
A: Hints:


*

*Think of  Euler's formulæ and rewrite 
$$\bigl(1+\cos\theta-i\sin\theta\bigr)^{n}=\bigl(1+\mathrm e^{-i\theta}\bigr)^{n}=\Bigl(\mathrm e^{-\tfrac{i\theta}2}\Bigl(\mathrm e^{\tfrac{i\theta}2}+\mathrm e^{-\tfrac{i\theta}2}\Bigr)\Bigr)^{\!n}=\mathrm e^{-\tfrac{ni\theta}2}\Bigl(2\cos\tfrac{i\theta}{2}\Bigr)^{\!n}.$$

*I would use the trigonometric formula
$$\cos3\theta=4\cos^3\theta-3\cos \theta$$
to deduce instantly the linearisation formula
$$\cos^3\theta=\frac14\bigl(\cos\theta+\cos 3\theta)$$
and transform the sum of cosines cubed into a linear combination of sums of cosines of arcs in arithmetic progression.

