Question

If 3 sin θ + 5 cos θ = 5, then write the value of 5 sin θ − 3 cos θ.

Answer 1

$$\begin{gathered} 3\sin \mathrm{\theta }+5\cos \mathrm{\theta }=5\left(\mathrm{Given}\right) \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ 9{\sin }^2\mathrm{\theta }+25{\cos }^2\mathrm{\theta }+30\sin \mathrm{\theta }\cos \mathrm{\theta }=25 \\ 30\sin \mathrm{\theta }\cos \mathrm{\theta }=25-9{\sin }^2\mathrm{\theta }-25{\cos }^2\mathrm{\theta }\left(1\right) \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}5\sin \mathrm{\theta }-3\cos \mathrm{\theta }. \\ {\left(5\sin \mathrm{\theta }-3\cos \mathrm{\theta }\right)}^2=25{\sin }^2\mathrm{\theta }+9{\cos }^2\mathrm{\theta }-30\sin \mathrm{\theta }\cos \mathrm{\theta } \\ {\left(5\sin \mathrm{\theta }-3\cos \mathrm{\theta }\right)}^2=25{\sin }^2\mathrm{\theta }+9{\cos }^2\mathrm{\theta }-\left(25-9{\sin }^2\mathrm{\theta }-25{\cos }^2\mathrm{\theta }\right)\left[\mathrm{From}\left(1\right)\right] \\ {\left(5\sin \mathrm{\theta }-3\cos \mathrm{\theta }\right)}^2=34{\sin }^2\mathrm{\theta }+34{\cos }^2\mathrm{\theta }-25 \\ {\left(5\sin \mathrm{\theta }-3\cos \mathrm{\theta }\right)}^2=34-25\left(\because {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\right) \\ {\left(5\sin \mathrm{\theta }-3\cos \mathrm{\theta }\right)}^2=9 \\ 5\sin \mathrm{\theta }-3\cos \mathrm{\theta }=\pm 3 \end{gathered}$$