Prove that sin10° +sin50° – sin70°= 0
Solution:
sin10° +sin50° – sin70°
= sin10° + 2 cos $\frac{50° + 70°}{2}$ sin $\frac{50° – 70°}{2}$
= sin10° + 2 cos $\frac{120°}{2}$ sin $ \left( – \frac{20°}{2}\right)$
= sin 10° – 2 cos 60° sin10°
= sin10° – 2 × $\frac{1}{2}$ sin10°
= sin10° – sin10°
= 0
∴ sin10° +sin50° – sin70°= 0 [Proved]