Question
Prove that:
$\sin40^\circ+\sin20^\circ=\cos10^\circ$

Answer

$\sin40^\circ+\sin20^\circ=\cos10^\circ$
$\text{LHS}=\sin40^\circ+\sin20^\circ$
$=\ 2\sin\Big(\frac{40^\circ+20^\circ}{2}\Big)\cos\Big(\frac{40^\circ-20^\circ}{2}\Big)$ $\Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$
$=\ 2\sin30^\circ\cos10^\circ$
$=\ 2\times\frac{1}{2}\cos10^\circ$
$=\ \cos10^\circ$
$=\ \text{RHS}$ $\Big[\because\ \sin30^\circ=\frac{1}{2}\Big]$

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