Question
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ prove that $\big(\text{m}^2+\text{n}^2\big)=\big(\text{a}^2+\text{b}^2\big).$
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Answer
We have $\text{m}^2+\text{n}^2=\big[(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2\big]$
$=\big(\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\cos\theta\sin\theta\big)\\ \ +\big(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta\big)$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta$
$=\big(\text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta\big)+\big(\text{b}^2\cos^2\theta+\text{b}^2\sin^2\theta\big)$
$=\text{a}^2\big(\cos^2\theta+\sin^2\theta\big)+\text{b}^2\big(\cos^2\theta+\sin^2\theta\big)$
$=\text{a}^2+\text{b}^2$ $\big[\because\ \sin^2+\cos^2=1\big]$
Hence, $\text{m}^2+\text{n}^2=\text{a}^2+\text{b}^2$
$=\big(\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\cos\theta\sin\theta\big)\\ \ +\big(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta\big)$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta$
$=\big(\text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta\big)+\big(\text{b}^2\cos^2\theta+\text{b}^2\sin^2\theta\big)$
$=\text{a}^2\big(\cos^2\theta+\sin^2\theta\big)+\text{b}^2\big(\cos^2\theta+\sin^2\theta\big)$
$=\text{a}^2+\text{b}^2$ $\big[\because\ \sin^2+\cos^2=1\big]$
Hence, $\text{m}^2+\text{n}^2=\text{a}^2+\text{b}^2$
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