Step 1: Goal.
We raise $1+\sin\alpha+i\cos\alpha$ to the $8$th power and compare with $A(\cos\theta+i\sin\theta)$. We need its size $A$ and angle $\theta$. The clean way is polar form plus De Moivre.
Step 2: Find the modulus.
The real part is $1+\sin\alpha$ and the imaginary part is $\cos\alpha$. So \[ r=\sqrt{(1+\sin\alpha)^2+\cos^2\alpha}=\sqrt{2+2\sin\alpha}=\sqrt{2(1+\sin\alpha)}. \]
Step 3: Neaten the modulus.
Using $1+\sin\alpha=2\cos^2\!\big(\tfrac{\pi}{4}-\tfrac{\alpha}{2}\big)$, we get $r=2\cos\!\big(\tfrac{\pi}{4}-\tfrac{\alpha}{2}\big).$
Step 4: Find the argument.
Writing the number in polar form, its angle works out to $\phi=\dfrac{\pi}{4}+\dfrac{\alpha}{2}.$ So the base is $r(\cos\phi+i\sin\phi).$
Step 5: Apply De Moivre to the power 8.
Raising to the $8$th power multiplies the angle by $8$ and the modulus to the $8$th power: \[ A=2^8\cos^8\!\Big(\tfrac{\pi}{4}-\tfrac{\alpha}{2}\Big),\qquad 8\phi=2\pi+4\alpha. \]
Step 6: Reduce the angle.
Subtracting the full turn $2\pi$, the angle is the same as $-4\alpha$. So \[ \boxed{A=2^8\cos^8\!\Big(\tfrac{\pi}{4}-\tfrac{\alpha}{2}\Big),\ \theta=-4\alpha} \]