Function Overlap

1. Here's a geometric approach: Consider a right angled triangle $ABC$ with its right angle at $C$ . Because the sum of the interior angles in a triangle is always $\pi$ , we have $\beta = \frac{\pi}{2} - \alpha$ . By definition, $\sin(\beta) = \frac{\overline{AC}}{\overline{AB}}$ and $\cos(\alpha)=\frac{\overline{AC}}{\overline{AB}}$ . Thus, $\cos(\alpha)=\sin(\beta)=\sin(\frac{\pi}{2} - \alpha) = \sin(\pi - (\frac{\pi}{2} - \alpha))=\sin(\alpha + \frac{\pi}{2})$ . If we swap the parameters, the signs change and hence $C=-\frac{\pi}{2}$ .

2. Alternatively, we may assume that one of the given answers is correct and use the addition formula for cosine: $\cos(x-\frac{\pi}{2})=\cos x \cdot \cos \left(-\frac{\pi}{2}\right) - \sin x \sin \left(\frac{\pi}{2}\right) = \cos x \cdot 0 - \sin x \cdot (-1) = \sin x$ will turn out to be the correct answer.

3. Using the complex definition of the cosine, we may also show this result using that $e^{i(-0.5\pi)}=\cos(-0.5\pi)+i\sin(-0.5\pi)=-i$ and $e^{i(0.5\pi)}=\cos(0.5\pi)+i\sin(0.5\pi)=i$ . We have

$\begin{aligned} \cos (x-0.5\pi) &= \frac{1}{2} \left( e^{i\cdot (-0.5\pi)} + e^{-i\cdot (-0.5\pi)} \right)\\ &= \frac{1}{2} \left( e^{i\cdot (x-0.5\pi)} + e^{-i\cdot (x-0.5\pi)} \right)\\ &= \frac{1}{2} \left( e^{i\cdot (x-0.5\pi)} + e^{i\cdot (0.5\pi-x))} \right)\\ &= \frac{1}{2} \left( e^{ix} \cdot e^{i\cdot (-0.5)} + e^{i\cdot 0.5}\cdot e^{-ix} \right)\\ &= \frac{1}{2} \left( -ie^{ix} + ie^{-ix} \right)\\ &= \frac{1}{2i} \left( -i\cdot i\cdot e^{ix} + i\cdot i\cdot e^{-ix} \right)\\ &= \frac{1}{2i} \left( e^{ix} - e^{-ix} \right)\\ &= \sin(x) \end{aligned}$