Question

If $\cos \left(\theta \right)+\sin \left(\theta \right)=\sqrt{2}\cos \left(\theta \right)$ prove that $\cos \left(\theta \right)-\sin \left(\theta \right)=\sqrt{2}\sin \left(\theta \right)$

Answer 1

Prove the given expression:

Given,

$$\begin{gathered} \cos \left(\theta \right)+\sin \left(\theta \right)=\sqrt{2}\cos \left(\theta \right) \\ \Rightarrow {\left(\cos \left(\theta \right)+\sin \left(\theta \right)\right)}^2={\left(\sqrt{2}\cos \left(\theta \right)\right)}^2\left[\mathrm{Squaring}\mathrm{on}\mathrm{both}\mathrm{side}\right] \\ \Rightarrow {\cos }^2\left(\theta \right)+2\cos \left(\theta \right)\sin \left(\theta \right)+{\sin }^2\left(\theta \right)=2{\cos }^2\left(\theta \right)\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right] \\ \Rightarrow 2\cos \left(\theta \right)\sin \left(\theta \right)=2{\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)-{\cos }^2\left(\theta \right) \\ \Rightarrow 2\cos \left(\theta \right)\sin \left(\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right) \\ \Rightarrow 2\cos \left(\theta \right)\sin \left(\theta \right)=\left(\cos \left(\theta \right)-\sin \left(\theta \right)\right)\left(\cos \left(\theta \right)+\sin \left(\theta \right)\right)\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ \Rightarrow 2\cos \left(\theta \right)\sin \left(\theta \right)=\left(\cos \left(\theta \right)-\sin \left(\theta \right)\right)\sqrt{2}\cos \left(\theta \right)\left[\because \cos \left(\theta \right)+\sin \left(\theta \right)=\sqrt{2}\cos \left(\theta \right)\right] \\ \Rightarrow \sqrt{2}\sin \left(\theta \right)=\left(\cos \left(\theta \right)-\sin \left(\theta \right)\right) \\ \therefore \cos \left(\theta \right)-\sin \left(\theta \right)=\sqrt{2}\sin \left(\theta \right) \end{gathered}$$

Hence proved, $\cos \left(\theta \right)-\sin \left(\theta \right)=\sqrt{2}\sin \left(\theta \right)$.