Question

sin(xy)+sin(yz)+sin(xz)=1

Answer 1

Answer:

[∵sin(x−z)=−sin(z−x)]

Step-by-step explanation:

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)]

=

2

1

siny[cosy−cos(2x−y)+cos(2z−y)−cosy]+

2

1

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

=

2

1

siny[cos(2z−y)−cos(2x−y)]+

2

1

sin(z−x)[cos(x−2y+z)−cos(z+x)]

=

2

1

siny[2sin(z+xy)sin(x−z)]+

2

1

sin(z−x)[2sin(z+x−y)siny]

=0

[∵sin(x−z)=−sin(z−x)].