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If $5 \cos x+12 \cos y=13$, then the maximum value of $5 \sin x+12 \sin y$ is :
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Verified Answer
The correct answer is:
$\sqrt{120}$
$\because 5 \cos x+12 \cos y=13$
$\Rightarrow \quad(5 \cos x+12 \cos y)^2=169$
$\therefore(5 \cos x+12 \cos y)^2+(5 \sin x+12 \sin y)^2$
$\Rightarrow \quad(13)^2+(5 \sin x+12 \sin y)^2$
$=25+144+120(\sin x \sin y+\cos x \cos y)$
$\Rightarrow \quad(5 \sin x+12 \sin y)^2=120 \cos (x-y)$
$\because \quad-1 \leq \cos (x-y) \leq 1$
$\therefore$ Maximum value of $5 \sin x+12 \sin y=\sqrt{120}$
$\Rightarrow \quad(5 \cos x+12 \cos y)^2=169$
$\therefore(5 \cos x+12 \cos y)^2+(5 \sin x+12 \sin y)^2$
$\Rightarrow \quad(13)^2+(5 \sin x+12 \sin y)^2$
$=25+144+120(\sin x \sin y+\cos x \cos y)$
$\Rightarrow \quad(5 \sin x+12 \sin y)^2=120 \cos (x-y)$
$\because \quad-1 \leq \cos (x-y) \leq 1$
$\therefore$ Maximum value of $5 \sin x+12 \sin y=\sqrt{120}$
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