Question
If $\text{x}=\text{a}\sin\theta+\text{b}\cos\theta$ and $\text{y}=\text{a}\cos\theta-\text{b}\sin\theta,$ prove that $\text{x}^2+\text{y}^2=\text{a}^2+\text{b}^2.$
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Answer
Given: $\text{x}=\text{a}\sin\theta+\text{b}\cos\theta$Squaring both sides, we get:
$\text{x}^2=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta\dots(\text{i})$
Also, $\text{y}=\text{a}\cos\theta-\text{b}\sin\theta$
Squaring both sides, we get:
$\text{y}^2=\text{a}^2\cos^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\dots(\text{ii})$
$\therefore\ \text{LHS}=\text{x}^2+\text{y}^2$
$=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\\+\text{a}^2\cos^2\theta-\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta$ [Using (i) and (ii)]
$=\text{a}^2\big(\sin^2\theta+\cos^2\theta\big)+\text{b}^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\text{a}^2+\text{b} ^2$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
Hence proved.
$\text{x}^2=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta\dots(\text{i})$
Also, $\text{y}=\text{a}\cos\theta-\text{b}\sin\theta$
Squaring both sides, we get:
$\text{y}^2=\text{a}^2\cos^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\dots(\text{ii})$
$\therefore\ \text{LHS}=\text{x}^2+\text{y}^2$
$=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\\+\text{a}^2\cos^2\theta-\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta$ [Using (i) and (ii)]
$=\text{a}^2\big(\sin^2\theta+\cos^2\theta\big)+\text{b}^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\text{a}^2+\text{b} ^2$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
Hence proved.
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