How do I show that $\cos(t)+1=2\cos^2(\frac{t}{2})$

Question

I have been trying to show that $\cos(t)+1$ is equal to $2\cos^2(\frac{t}{2})$. Here's what I have done so far: $$\cos(t)+1 $$ Using the identity: $$1=2\cos^2(t)-\cos(2t)$$ $$\cos(t)+(2\cos^2(t)-\cos(2t))$$ $$\cos(t)-\cos(2t)+2\cos^2(t)$$ $\cos(2t)$ can be written as $\cos^2(t)-\sin^2(t)$ $$\cos(t)-(\cos^2(t)-\sin^2(t))+2\cos^2(t) $$ $$\cos(t)-\cos^2(t)+2\cos^2(t)+\sin^2(t)$$ $$\cos(t)+\cos^2(t)+\sin^2(t) $$ $$\cos(t)+1$$ I keep returning to what I started with. Need help.

Answer 1

Notice that

$$ \cos ( 2 \alpha ) = \cos^2 \alpha - \sin^2 \alpha $$

Therefore, with $\alpha = t/2$, one has

$$ \cos ( t ) = \cos^2(t/2) - \sin^2(t/2) $$

since $\sin^2 (t/2) = 1 - \cos^2 (t/2)$ , one has the result.

Answer 2

Let $f(t) = 1+\cos t - 2 \cos^2 {t \over 2}$ and note that $f(0) = 0$, $f'(0) = 0$ and $f''= -f$.

Answer 3

Hint: if you know the half-angle formula $$\cos \left(\frac t 2\right ) = \sqrt \frac {\cos t+1}{2}$$ then $$\cos^2 \left(\frac t 2\right ) = \ ?$$