Question

If $x=3sin\theta andy=4cos\theta ,$ find the value of $\sqrt{16x^2+9y^2}.$

Answer 1

$x=3sin\theta$
Squaring both sides
$\Rightarrow$ $x^2=9si{n}^2\theta$
$si{n}^2\theta =\frac{x^2}{9}$ ...(i)
And $y=4cos\theta$
Squaring both sides
$y^2=16co{s}^2\theta$
$co{s}^2\theta =\frac{y^2}{16}$ ...(ii)
On adding eq. (i) and eq. (ii)
$si{n}^2\theta +co{s}^2\theta =\frac{x^2}{9}+\frac{y^2}{16}$
$1=\frac{x^2}{9}+\frac{y^2}{16}$ (Since, $si{n}^2\theta +co{s}^2\theta =1$)
$1=\frac{16x^2+9y^2}{144}$
$16x^2+9y^2=144$
$\sqrt{16x^2+9y^2}=\sqrt{144}$
$\sqrt{16x^2+9y^2}=12$