If x + y =pi + z, then prove that sin²x + sin²y – sin²z = 2 sin x sin y cos z.
Answers
Given :
- x + y = π + z
To prove :
- sin²x + sin²y - sin²z = 2 sin x sin y sin z
Proof :
As given
→ x + y = π + z
→ sin ( x + y ) = sin ( π + z )
putting sin ( π + z ) = - sin z
→ sin ( x + y ) = - sin z
using trigonometric identity
sin ( A + B ) = sin A cos B + cos A sin B
→ sin x cos y +cos x sin y = - sin z
squaring both sides
and using idnetity (a + b)² = a² + b² + 2 ab in LHS
→ sin²x cos²y +cos²x sin²y + 2 sin x sin y cos x cos y = sin²z
putting cos²y = 1 - sin²y and putting cos²x = ( 1 - sin² x )
→ sin² x ( 1 - sin²y ) + ( 1 - sin²x ) sin²y + 2 sin x sin y cos x cos y = sin²z
→ sin²x - sin²x sin²y + cos²x - sin²x sin²y + 2 sin x sin y cos x cos y = sin²z
→ sin²x + sin²y - sin²z = 2 sin²x sin²y - 2 sin x sin y cos x cos y
taking ( sin x sin y ) common in RHS
→ sin²x + sin²y - sin²z = 2 sin x sin y ( sin x sin y - cos x cos y )
Using trigonometric identity, putting ( sin x sin y - cos x cos y ) = cos ( x + y ) in RHS
→ sin²x + sin²y - sin²z = 2 sin x sin y cos ( x + y )
As given that (x + y) = (π + z)
→ sin²x + sin²y - sin²z = 2 sin x sin y cos ( π + z )
since, by identity cos ( π + z ) = cos z
therefore,
→ sin²x + sin²y - sin²z = 2 sin x sin y cos z
PROVED .