Question
Prove that:
$\sin51^\circ+\cos81^\circ=\cos21^\circ$

Answer

Consider LHS:
$\sin51^\circ+\cos81^\circ$
$=\ \sin51^\circ+\cos(90^\circ-9^\circ)$
$=\ \sin51^\circ+\sin9^\circ$
$=\ 2\sin\Big(\frac{51^\circ+9^\circ}{2}\Big)\cos\Big(\frac{51^\circ-9^\circ}{2}\Big)$ $\Big\{\because\ \sin\text{A}+\sin\text{B}=-2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big\}$
$=\ 2\sin30^\circ\cos21^\circ$
$=\ 2\times\frac{1}{2}\cos(21^\circ)$
$=\ \sin(21^\circ)$
$=\ \text{RHS}$
Hence, LHS = RHS.

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