Question

lf $5\cos x+12\cos y=13$, then the maximum value of $5\sin x+12\sin y$ is

• A. $12$
• B. $\sqrt{120}$
• C. $\sqrt{20}$
• D. $13$

Answer 1

The correct option is B $\sqrt{120}$

Given

$(5\cos x+12\cos y)=13$

Squaring both side

$(5\cos x+12\cos y{)}^2={13}^2=169$

$25{\cos }^{2}x+120\cos x\cos y+144{\cos }^{2}y=\mathrm{169....................}\left(1\right)$

Suppose, $5\sin x+12\sin y=A$

$25{\sin }^{2}x+120\sin x\sin y+144{\sin }^{2}y=A^2.............\left(2\right)$

Adding (1)+(2) we get

$25+144+120(\cos x\cos y+\sin x\sin y)=A^2+169$

$169+120(\cos x\cos y+\sin x\sin y)=A^2+169$

$120(\cos x\cos y+\sin x\sin y)=A^2$

$120\cos (x-y)=A^2$

$A^2$ is max when $\cos (x-y)=1$

so,

$A_{max}^2=120$

$A_{max}=\sqrt{120}$

$5\sin x+12\sin y=A_{max}=\sqrt{120}$