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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...