Prove $\sin(45^°) + \sin(15^°) = \sin(75^°)$

Question

I rewrote the statement as $$ \sin(30^° + 15^°) + \sin(15^°) = \cos(15^°). $$ Then I got $$ (\sqrt{3}-2) \sin(15^°) = \cos(15^°). $$

Answer 1

$$\sin 75^° - \sin 15^° = 2\sin\left(\frac{75^°-15^°}{2}\right)\cos\left(\frac{75^°+15^°}{2}\right) = 2 \sin 30^° \cos 45^° $$$$= \cos 45 ^°=\sin 45^°.$$

Answer 2

\begin{align} \sin(45^{°}) + \sin(15^{°}) & = 2 \sin\left(\dfrac{45^{°}+15^{°}}2\right) \cos\left(\dfrac{45^{°}-15^{°}}2\right) = 2 \sin(30^{°})\cos(15^{°})\\ & = 2 \cdot \dfrac12 \cdot \sin(90^{°}-15^{°}) = \sin(75^{°}) \end{align}