Question

sin50°-sin70°+sin10°=0

Answer 1

Use the sum formula

$\sin A- \sin B=2\sin(\frac{A-B}{2}) \cos(\frac{A+B}{2})$

Now using the above formula,

$\sin 70^o- \sin 10^o=2\sin(\frac{70^o-10^o}{2}) \cos(\frac{70^o+10^o}{2})\\ \sin 70^o- \sin 10^o=2\sin(30^o) \cos(40^o)\\ \sin 70^o- \sin 10^o=2(\frac{1}{2}) \cos(90^o-50^o) \\ \sin 70^o- \sin 10^o= \cos(90^o-50^o) \\ \sin 70^o- \sin 10^o= \sin(50^o)$

Thus,

$LHS= \sin(50^o) -(\sin 70^o- \sin 10^o)=\sin(50^o) - \sin(50^o) =0=RHS$

The proof is complete.

Answer 2

Answer

Step-by-step explanation:

sin50 - sin70 +sin10 = 0

sin(60-10) - sin(60+10) + sin 10 =0

sin60cos10 - cos60sin10 - sin60cos10 - cos60sin10 +sin10 = 0

$\frac{\sqrt{3} }{2}$cos10 - $\frac{1}{2\\}$sin 10 - $\frac{\sqrt{3} }{2}$cos 10 - $\frac{1}{2}$sin10 +sin10 =0

sin10 -sin10 =0

0=0

Hence proved