Prove the identity $1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$ Here is the problem:
                     $$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
Can you help me prove  that this is an trigonometric identity?
 A: $$
1+\sin x = 1 + \cos (90^o-x) = 2 \cos^2 \left ( \frac {90^o - x}2\right ) = 2 \cos^2 \left ( 45^o -\frac x2\right )
$$
A: As $\cos 2y=2\cos^2y-1$
$\sin x=\cos(90^\circ-x)=\cos 2\left(45^\circ-\frac x2\right)=2\cos^2(45^\circ-\frac x2)-1$
A: $$\begin{align}
2\cdot \color{blue}{\cos^2\left(45^{\circ}-\frac{x}{2}\right)}&=\\
2\cdot \color{blue}{\frac{1+cos(double angle)}{2}}&=\\
2\cdot \color{blue}{\frac{1+cos(90^{\circ} - x)}{2}}&=\\
1+\cos\left(90^{\circ}-x\right) &= \\
1+\sin(x)&\\
\end{align}$$
A: $$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
$$\cos{2\alpha}=2cos^{2}{\alpha}-1$$
$$\cos{2\cdot \left(45^{\circ}-\frac{x}{2}\right)}=\cos{\left(90^{\circ}-x\right)}=\cos{90^{\circ}\cos{x}}+\sin{90^{\circ}\sin{x}}=0+1\cdot \sin{x}=\sin{x}.$$
A: It always annoyed (or rather bored) me to play with trigonometric identities. It was great to find out about the connection with exponents that reduces trigonometry to easy power calculations. Here is a solution via exponents:
The basic identities are
$$
\begin{array}{lll}
e^{ix}&=&\cos x + i \sin x\\
\cos x&=& \frac{e^{ix}+e^{-ix}}{2}\\
\sin x&=& \frac{e^{ix}-e^{-ix}}{2i}.
\end{array}
$$
BUT here $x$ must be in radians, so $45^\circ$ becomes $\pi/4$. 
The terms in the identity then become
$$
\begin{array}{rcl}
1+\sin x &=& 1+\frac{e^{ix}-e^{-ix}}{2i}\quad\mbox{and}\\
2(\cos(45^\circ-x/2))^2 &=& 2\left(\frac{e^{i(\pi/4-x/2)}+e^{-i(\pi/4-x/2)}}{2}\right)^2\\
&=&\frac{1}{2}\big(e^{i2(\pi/4-x/2)}+2e^{i(\pi/4-x/2)-i(\pi/4-x/2)}+e^{-i2(\pi/4-x/2)}\big)\\
&=&\frac{1}{2}\big(e^{i\pi/2}e^{-ix}+2e^{0}+e^{-i\pi/2}e^{-ix}\big)=\cdots
\end{array}
$$
Here we used the usual power rules: $(a^{b})^c=a^{bc}$ and $a^{b+c}=a^ba^c$.
Now plug in $e^{0}=1$, $e^{i\pi/2}=\cos 90^\circ + i\sin90^\circ=i$, $e^{-i\pi/2}=-i$ and $1/i=-i$: 
$$
\cdots = \frac{1}{2}(2+ ie^{-ix}-ie^{ix})=1+\frac{e^{ix}-e^{-ix}}{2i}=1+\sin x,
$$
as required.
