Question

If $5\sin \theta +3\cos \theta =4$, find the value of $3\sin \theta -5\cos \theta$.

Answer 1

Consider the given equation.

$5\sin \theta +3\cos \theta =4$ …… (1)

On squaring both sides, we get

$25{\sin }^2\theta +9{\cos }^2\theta +30\sin \theta \cos \theta =16$

$16{\sin }^2\theta +9({\sin }^2\theta +{\cos }^2\theta )+30\sin \theta \cos \theta =16$

$16{\sin }^2\theta +30\sin \theta \cos \theta =16-9=7$ ……. (2)

Suppose that,

$3\sin \theta -5\cos \theta =p$

On squaring both sides, we get

$9{\sin }^2\theta +25{\cos }^2\theta -30\sin \theta \cos \theta =p^2$ ……. (3)

On adding equation (2) and (3), we get

$9{\sin }^2\theta +25{\cos }^2\theta +16{\sin }^2\theta =7+p^2$

$25{\sin }^2\theta +25{\cos }^2\theta =7+p^2$

$7+p^2=25({\sin }^2\theta +{\cos }^2\theta )$

$p^2=25-7$

$p^2=18$

$p=\pm 3\sqrt{2}$

Hence, the value of $3\sin \theta -5\cos \theta$ is $\pm 3\sqrt{2}$.