Question

Transformation of sums and product proof
$\sin(10 ) + \sin(50 ) - \sin(70) = 0$

Answer 1

$\sin(c) - \sin(d) = 2 \times \cos( \frac{c + d}{2} ) \times \sin \frac{(c - d)}{2}$
sin 50 - sin 70= 2 cos ((50+70)/2) sin ((50-70)/2)
                          = 2 cos (120/2) sin(-20/2) 
                          = 2 cos 60 sin (-10)
sin 50 - sin 70= -sin 10       [cos 60= 1/2 and  sin(-A)= -sin A]
Therefore, sin 50 - sin 70 + sin 10= -sin 10 + sin 10= 0

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