Question

Prove that sin 50° - sin 70° + sin 10° = 0.

Answer 1

Answer:

Step-by-step explanation:

As we know that

$sin\:C-sin\:D=2\:cos\frac{(C+D)}{2} \:sin\frac{(C-D)}{2} \\\\\\sin\:50-sin\:70=2\:cos\frac{(50+70)}{2} \:sin\frac{(50-70)}{2}\\\\\\=2\:cos\:60\:\:sin(-10)\\\\sin\:\:50- sin\:\:70 + sin\:\:10=2\:cos\:60\:\:sin(-10)+sin\:10\\\\\\cos\:60=\frac{1}{2}\\\\sin(-\theta)=-sin(\theta) \\\\=-2\times\frac{1}{2} sin\:10+sin\:10\\\\=-sin\:10+sin\:10\\\\=0$

hence proved