Question

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

Answer 1

Let $x=\sin {10}^0+\sin {20}^0+\sin {40}^0+\sin {50}^0=\sin {10}^0+\sin {50}^0+\sin {20}^0+\sin {40}^0$

Applying trigonometric identity

$\sin a^0+\sin b^0$ $=2\sin \left(\frac{a^0+b^0}{2}\right)\cos \left(\frac{a^0-b^0}{2}\right)$

$x==2\sin \left(\frac{{10}^0+{50}^0}{2}\right)\cos \left(\frac{{20}^0-{50}^0}{2}\right)+2\sin \left(\frac{{20}^0+{40}^0}{2}\right)\cos \left(\frac{{20}^0-{40}^0}{2}\right)$

$=2\sin {30}^0\cos (-{20}^0)+2\sin {30}^0\cos (-{10}^0)$

$=2\times \frac{1}{2}\cos {20}^0+2\times \frac{1}{2}\cos {10}^0=\cos {20}^0+\cos {10}^0$

$=\cos ({90}^0-{20}^0)+\cos ({90}^0-{10}^0)=\sin {70}^0+\sin {80}^0$

$\therefore$ $\sin {10}^0+\sin {20}^0+\sin {40}^0+\sin {50}^0=\sin {70}^0+\sin {80}^0$

Hence Proved.