trig identities question (1 Viewer)

[Aviator_13]

Prove this question:

sin ( θ + α ) sin ( θ − α ) = sin² θ − sin² α

 

[fan96]

Use the identity:

 

[potat_o]

Using sin(A+B) = (sin A) cos(B) + cos(A) sin(B) and sin(A-B) = (sin A) cos(B) - cos(A) sin(B)

To prove sin ( θ + α ) sin ( θ − α ) = sin^2θ − sin^2a
Taking the LHS,

sin ( θ + α ) sin ( θ − α )
= [sin(θ)cos(a)+cos(θ)sin(a)] [sin(θ)cos(a)-cos(θ)sin(θ)]
=sin^2θcos^2a-cos^2θsin^2a (Expanding and cancelling)
= sin^2θ(1-sin^2a)-(1-sin^2θ)sin^2a (Using sin^2θ+cos^θ = 1)
=sin^2θ-sin^2θsin^2a-[sin^2a-sin^2θsin^2a)
=sin^2θ-sin^2θsin^2a-sin^2a+sin^2θsin^2a)
= sin^2θ - sin^2a=RHS