Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?
 A: Yes, 
$$\cos^2\theta - \sin^2\theta = (\cos\theta)(\cos\theta) - (\sin\theta)(\sin\theta)$$
$$=\cos(\theta + \theta) = \cos(2\theta)$$
A: Yes, indeed! Since the sine and cosine functions are real-valued functions on the reals, then since $$\cos2\theta+i\sin 2\theta=\cos^2\theta-\sin^2\theta+i(2\sin\theta\cos\theta),$$ we have: $$\cos2\theta=\cos^2\theta-\sin^2\theta\\\sin 2\theta=2\sin\theta\cos\theta$$
A: using de moivres theorem we can generalise.$$\cos(n\theta)=\cos^n(\theta)-\frac{n(n-1)}{2!}\cos^{n-2}(\theta)\sin^2(\theta)+\frac{n(n-1)(n-2)(n-3)}{4!} \cos^{n-4}(\theta)\sin^4(\theta)........$$
A: Here is a geometric proof:

*

*the triangle ABC is isosceles $AB = AC = r$ and equiangular with equal angles at B and C

*the angle at A being $2\theta$, both angles at B and C equal $\pi/2-\theta$

*the height = perpendicular bisector from A to BC meets the latter at D

*the height perpendicular from C to AB meets the latter at H

*now express the length of AH first as cathetus of rectangular triangle CAH

*and then as difference of AB and HB. HB is a cathetus in rectangular triangle BCH.
Putting this together we get
$$\cos2\theta = 1-2\sin^2\theta$$
Using $\cos^2\theta+\sin^2\theta=1$ (Pythagoras) the equivalent expressions
$$\cos2\theta = \cos^2\theta-\sin^2\theta=2\cos^2\theta-1$$ are as well obtained.


