Question

16. \( \cos \left\{\theta+\left(n-\frac{3}{2}\right) \phi\right\}-\cos \left\{\theta+\left(n+\frac{3}{2}\right) \phi\right\}=2 \sin \frac{3 \phi}{2} \cdot \sin (\theta+n \phi) \)

Answer 1

cos(a-b) - cos (a+b) = 2 sina sinb

x=theta, y=phi

LHS=cos[(x+yn)-1.5y)]-cos[(x+yn)+1.5y)]

accoring to formula let a= x+yn,b=1.5 y

Therefore , LHS= 2 sin(x+yn) sin (1.5y)

=2sin (3y/2)sin(theta + n*phi) = RHS