Question

that $\sin {10}^{\circ }+\sin {20}^{\circ }+\sin {40}^{\circ }+\sin 5{\theta }^{\circ }=\sin {70}^{\circ }+\sin {80}^{\circ }$

Answer 1

$LHS=\sin {10}^{\circ }+\sin {20}^{\circ }+\sin {40}^{\circ }+\sin {50}^{\circ }$

$=(\sin {10}^{\circ }+\sin {50}^{\circ })+(\sin {20}^{\circ }+\sin {40}^{\circ })$

$=2\sin {30}^{\circ }\cos {20}^{\circ }+2\sin {30}^{\circ }\cos {10}^{\circ }$

$=\cos {20}^{\circ }+\cos {10}^{\circ }=\sin {70}^{\circ }+\sin {80}^{\circ }=RHS$

$LHS=RHS$

Hence proved