Aviator_13
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- 2019
Prove this question:
sin ( θ + α ) sin ( θ − α ) = sin2 θ − sin2 α
sin ( θ + α ) sin ( θ − α ) = sin2 θ − sin2 α
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trig identities question (1 Viewer)
- Thread starter Aviator_13
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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
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