Question
Prove that:
$\sin23^\circ+\sin37^\circ=\cos7^\circ$

Answer

$\sin23^\circ+\sin37^\circ=\cos7^\circ$
$\text{LHS}=\sin23^\circ+\sin37^\circ$
$=\ 2\sin\Big(\frac{23^\circ+37^\circ}{2}\Big)\cos\Big(\frac{23^\circ-37^\circ}{2}\Big)$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin(30^\circ)\cos(-7^\circ)$
$=\ 2\times\frac{1}{2}\cos7^\circ\Big[\because\ \cos(-\theta)=\cos\theta,\sin30^\circ=\frac{1}{2}\Big]$
$=\ \cos7^\circ=\text{RHS}$

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