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Question
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that (m2 + n2) = (a2 + b2).
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Solution
We have `m^2 + n^2 = [(a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2]`
= `(a^2 cos^2 theta + b^2 sin ^2 theta + 2 ab cos theta sin theta) + (a^2 sin^2 theta + b^2 cos^2 theta -2ab cos theta sin theta)`
= `a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 cos^2 theta`
= `(a^2 cos^2 theta + b^2 sin^2 theta) + (b^2 cos^2 theta + b^2 sin^2 theta)`
= `a^2 (cos^2 theta + sin^2 theta ) + b^2 (cos^2 theta + sin^2 theta)`
= `a^2 + b^2 [∵ sin^2 + cos^2 = 1]`
Hence, `m^2 + n^2 = a^2 + b^2`
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