Question
Prove that: $\sin50^\circ+\sin10^\circ=\cos20^\circ$

Answer

$\sin50^\circ+\sin10^\circ=\cos20^\circ$ $\text{LHS}=\sin50^\circ+\sin10^\circ\ \Big[\because\ \sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}\Big]$ $\sin50^\circ+\sin10^\circ=2\sin\frac{60^\circ}{2}\cos20^\circ$ $=\ 2\sin30^\circ\cos20^\circ$ $=\ 2\times\frac{1}{2}\cos20^\circ$ $=\ \cos20^\circ=\text{RHS}\ \Big[\because30^\circ=\frac{1}{2}\Big]$

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