Prove that
$sin 10^{\circ} + sin 20^{\circ} + sin 40^{\circ} + sin 50^{\circ} = sin 70^{\circ} + sin 80^{\circ}$ .

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Solution

$sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$

LHS: $sin 10 + sin 20 + sin 40 + sin 50$

$\Rightarrow (sin 50 + sin 10) + (sin 40 + sin 20)$

$\Rightarrow (2 sin (\frac{50 + 10}{2}) cos (\frac{50 - 10}{2})) + (2 sin (\frac{40 + 20}{2}) cos (\frac{40 - 20}{2}))$

$\Rightarrow 2 sin 30 cos 20 + 2 sin 30 cos 10$

$\Rightarrow 2 sin 30 (cos 20 + cos 10)$

$\Rightarrow 2 \times \frac{1}{2} (cos 20 + cos 10) \Rightarrow cos A + cos B$

$cos (90 - θ) = sin θ$

$\Rightarrow cos 20 + cos 10$

$\Rightarrow cos (90 - 70) + cos (90 - 80)$

$\Rightarrow sin 70 + sin 80$

Hence proved.

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