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 $\left( – \frac{22°}{2}\right)$ +sin83°
= 2 sin30° cos11°+sin83°
= 2×$\frac{1}{2}$ coc 11° +sin83°
= cos11°+sin(90°-7°)
=cos11°+cos7°
R.H.S: sin23°+sin37°+sin79°
=2sin $\frac{23° + 37°}{2}$ cos $\frac{23° – 37°}{2}$ +sin79°
= 2sin $\frac{60°}{2}$ cos $\left( – \frac{14°}{2}\right)$ +sin79°
= 2sin30° cos7°+sin79°
=2×$\frac{1}{2}$×cos7°+sin79°
= cos7° +sin79°
= cos7°+sin(90°-11°)
= cos7°+cos11°
∴ L.H.S=R.H.S [Proved]