If (a cos θ - b sin θ) = c, show that: (a sin θ + b cos θ) =
.
Consider: (a cos θ - b sin θ) 2 + (a sin θ + b cos θ) 2
= a 2 cos 2 θ + b 2 sin 2 θ – 2ab sin θ cos θ + a 2 sin 2 θ + b 2 cos 2 θ + 2ab sin θ cos θ
= a 2 (cos 2 θ + sin 2 θ ) + b 2 (sin 2 θ + cos 2 θ ) – 2ab sin θ cos θ + 2ab sin θ cos θ
= a 2 + b 2 ( ∵ sin 2 θ + cos 2 θ = 1)
∴ (a cos θ - b sin θ) 2 + (a sin θ + b cos θ) 2 = a 2 + b 2
∴ (a cos θ - b sin θ) 2 = a 2 + b 2 - (a sin θ + b cos θ) 2
= a 2 + b 2 - c 2 (from equation (1))
∴ (a sin θ + b cos θ) = 
.
Hence, proved.
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