where does the $\cos(\theta)=1-2\sin^{2}(\theta/2)$ come from? I have seen people using $\cos(\theta)=1-2\sin^{2}(\theta/2)$ but how do we get it?
 A: It's a special case of the compound-angle formula $\cos (A+B)=\cos A\cos B-\sin A\sin B$. Take $A=B=\frac{\theta}{2}$ so $\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$. This can be written in two equivalent forms using $\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}=1$, one being $1-2\sin^2\frac{\theta}{2}$ (the other is $2\cos^2\frac{\theta}{2}-1$).
A: $$
\cos(2\theta)=1-2\sin^2(\theta).
$$
Change $\theta\rightarrow\theta/2$.
A: Hint: What happens if you use the double-angle identity for cosine and use the substitution $\theta \to \frac{\theta}{2}$?
Also, note that
$$\cos(a\pm b) = \cos a\cos b\mp\sin a\sin b$$
which allows you to derive the double-angle identity.
A: Hint: Use the definition of the half angle formula $$\sin \frac {\theta} 2 = \pm \sqrt {\frac {1 + \cos \theta}{2}}$$ Square both sides and manipulate as needed.
A: De Moivre:$$\cos^2\alpha-\sin^2\alpha+2i\cos\alpha\sin\alpha=(\cos\alpha+i\sin\alpha)^2=\cos(2\alpha)+2i\sin(2\alpha)$$
So that: $$1-2\sin^2\alpha=\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha=\cos^2\alpha-\sin^2\alpha=\cos2\alpha$$
Now substitute: $\alpha=\frac12\theta$.
