If a cos θ + b sin θ = m and a sin θ − b cos θ


If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m2 + n2) = (a2 + b2).


We have $m^{2}+n^{2}=\left[(a \cos \theta+b \sin \theta)^{2}+(a \sin \theta-b \cos \theta)^{2}\right]$

$=\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+2 a b \cos \theta \sin \theta\right)+\left(a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta\right)$

$=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta$

$=\left(a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+\left(b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\right)$

$=a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+b^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$=a^{2}+b^{2} \quad\left[\because \sin ^{2}+\cos ^{2}=1\right]$

Hence, $\mathrm{m}^{2}+n^{2}=a^{2}+b^{2}$

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