Express a complex number in modulus amplitude form
$\displaystyle 1+\sin \alpha +i\cos \alpha $
My Attempt:
$\displaystyle r\cos \theta= 1+\sin \alpha $
$\displaystyle r\sin \theta= \cos \alpha $
Squaring and adding.. $\displaystyle r^2= (1+\sin \alpha)^2+ \cos^2 \alpha$
$\displaystyle r^2= 2(1+\sin \alpha) $
$\displaystyle \tan \theta = \frac{\cos \alpha}{1+\sin \alpha} $
How to break up $\displaystyle 1+\sin \alpha $?
It is better to solve it this way:
$$\begin{aligned} 1+\sin\alpha+i\cos \alpha &=1+\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)\\ & \stackrel{*}{=}2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\\ &=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\left(\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+i\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right)\\ &=2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)e^{i\left(\pi/4-\alpha/2\right)}\\ \end{aligned}$$
$(*)$, In this step I used the following formulas:
$\cos(2x)=2\cos^2x-1$ and $\sin(2x)=2\sin x\cos x$
In your method, I am not seeing how you get that expression for $\tan\theta$. Rather it should be $\tan\theta=\dfrac{\cos\alpha}{1+\sin\alpha}$. Hence,
$$\tan\theta=\frac{\sin\left(\frac{\pi}{2}-\alpha\right)}{1+\cos\left(\frac{\pi}{2}-\alpha\right)}=\frac{2\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$$ $$\Rightarrow \tan\theta=\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$$
Hint
Another way would consist in writing $$\displaystyle 1+\sin( \alpha) +i\cos( \alpha)=\Big(\sin(\frac{\pi}{2})+\sin( \alpha)\Big)+i \Big( \cos(\frac{\pi}{2})+\cos( \alpha)\Big)$$ Now, apply $$\sin(p)+\sin(q)=2 \sin \left(\frac{p+q}{2}\right) \cos \left(\frac{p-q}{2}\right)$$ $$\cos(p)+\cos(q)=2 \cos \left(\frac{p+q}{2}\right) \cos \left(\frac{p-q}{2}\right)$$
I am sure that you can take from here.
A general fact is: $$\forall z\in\mathbb{C},\ \lvert z\rvert^2+z^2=z(\overline z+z)=2\Re(z)z.$$ (a manifestation of the inscribed angle theorem?). Hence: $$\forall z\in\mathbb{C},\ \bigl\lvert\lvert z\rvert^2+z^2\bigr\rvert=2\bigl\lvert\Re(z)\bigr\rvert\lvert z\rvert,\ \text{and}\ \arg(z)=\begin{cases}\arg(z)&\text{if $\Re(z)>0$}\\\arg(z)+\pi&\text{if $\Re(z)<0$}\end{cases}$$ In your case, setting: $$z=\mathrm{e}^{i(-\alpha/2+\pi/4)}$$ yields $$z^2=\mathrm{e}^{i(-\alpha+\pi/2)}=i\mathrm{e}^{-i\alpha}=i(\cos\alpha-i\sin\alpha)=\sin\alpha+i\cos\alpha$$ and $$\lvert z\rvert=1.$$ We also have $$\Re(z)=\cos(-\alpha/2+\pi/4).$$ Hence $$1+\sin\alpha+i\cos\alpha=\begin{cases}2\cos(-\alpha/2+\pi/4)\mathrm{e}^{i(-\alpha/2+\pi/4)}&\text{if $\cos(-\alpha/2+\pi/4)\geq0$}\\-2\cos(-\alpha/2+\pi/4)\mathrm{e}^{i(-\alpha/2-3\pi/4)}&\text{if $\cos(-\alpha/2+\pi/4)\leq0$.}\end{cases}$$
