Prove that, sin19°+sin41°+sin83°=sin23°+sin37°+sin79°
Prove that, sin19°+sin41°+sin83°=sin23°+sin37°+sin79° Solution: L.H.S: sin19°+sin41°+sin83° = 2 sin$\frac{19° + 41°}{2}$ cos $\frac{19° -41°}{2}$ +sin83° = 2 sin $\left(\frac{60°}{2}\right)$ cos …