Question

30. Prove that:
(i) sin (x + y) sin (x - y) + sin (y + z) sin (y - z) + sin (z + x) sin (z - x) = 0
​

Answer 1

Answer:

Combining the 1st two and last two terms, we get

L.H.S. siny[sinxsin(x−y)+sinzsin(y−z)]+sin(z−x)[sinzsinx+sin(x−y)sin(y−z)]

=21siny[cosy−cos(2x−y)+cos(2z−y)−cosy]+21sin(z−x)[cos(z−x)−cos(z+x)+cos(x−2y+z)−cos(x−z)]

=21siny[cos(2z−y)−cos(2x−y)]+21sin(z−x)[cos(x−2y+z)−cos(z+x)]

=21siny[2sin(z+xy)sin(x−z)]+21sin(z−x)[2sin(z+x−y)siny]=0

[∵sin(x−z)=−sin(z−x)].

Step-by-step explanation:

I hope that '& help you...