Question

Given that $\sin \left(\mathrm{\theta }\right)+2\cos \left(\mathrm{\theta }\right)=1$, then prove that $2\sin \left(\mathrm{\theta }\right)-\cos \left(\mathrm{\theta }\right)=2$.

Answer 1

Show that $2\sin \theta -\cos \theta =2$

Squaring both the sides

$$\begin{gathered} {(\sin \theta +2\cos \theta )}^2=1^2 \\ \Rightarrow {\sin }^2\theta +4{\cos }^2\theta +4\sin \theta \cos \theta =1 \\ \Rightarrow (1-{\cos }^2\theta )+4(1-{\sin }^2\theta )+4\sin \theta \cos \theta =1[{\sin }^2\theta +{\cos }^2\mathrm{\theta }=1] \\ \Rightarrow 4{\sin }^2\theta +{\cos }^2\theta -4\sin \theta \cos \theta =4[a^2+b^2+2ab={(a-b)}^2] \\ \Rightarrow {(2\sin \theta -\cos \theta )}^2=4 \\ \Rightarrow 2sin\theta -cos\theta =2 \end{gathered}$$

Hence Proved $2sin\theta -cos\theta =2$.