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If $\cos \theta +\sin \theta =\sqrt{2}\cos \theta $, show that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta $.

Last updated date: 17th Jul 2024
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Hint: We first take the square value of the equation $\cos \theta +\sin \theta =\sqrt{2}\cos \theta $. We then use the identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to break the expression. Putting the value of $\cos \theta +\sin \theta =\sqrt{2}\cos \theta $, we prove that $\cos \theta -\sin \theta =\sqrt{2}\sin \theta $.

Complete step-by-step solution:
It is given that $\cos \theta +\sin \theta =\sqrt{2}\cos \theta $.
We take square values on both sides of the equation.
So, we get ${{\left( \cos \theta +\sin \theta \right)}^{2}}={{\left( \sqrt{2}\cos \theta \right)}^{2}}$. We sue the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Simplifying we get
  & {{\left( \cos \theta +\sin \theta \right)}^{2}}={{\left( \sqrt{2}\cos \theta \right)}^{2}} \\
 & \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta =2{{\cos }^{2}}\theta \\
 & \Rightarrow {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2\cos \theta \sin \theta \\
Now we use the factorisation identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We get ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\left( \cos \theta +\sin \theta \right)\left( \cos \theta -\sin \theta \right)$.
So, we get $\left( \cos \theta +\sin \theta \right)\left( \cos \theta -\sin \theta \right)=2\cos \theta \sin \theta $.
We place the values given of $\cos \theta +\sin \theta =\sqrt{2}\cos \theta $ and get
  & \left( \cos \theta +\sin \theta \right)\left( \cos \theta -\sin \theta \right)=2\cos \theta \sin \theta \\
 & \Rightarrow \left( \sqrt{2}\cos \theta \right)\left( \cos \theta -\sin \theta \right)=2\cos \theta \sin \theta \\
Dividing both sides with $\sqrt{2}\cos \theta $, we get
  & \left( \sqrt{2}\cos \theta \right)\left( \cos \theta -\sin \theta \right)=2\cos \theta \sin \theta \\
 & \Rightarrow \cos \theta -\sin \theta =\dfrac{2\cos \theta \sin \theta }{\sqrt{2}\cos \theta }=\sqrt{2}\sin \theta \\
Thus, proved $\cos \theta -\sin \theta =\sqrt{2}\sin \theta $.

Note: It is important to remember that the condition to eliminate the $\cos \theta $ from both denominator and numerator is $\cos \theta \ne 0$. No domain is given for the variable $x$. The value of $\cos \theta \ne 0$ is essential. The simplified condition will be $x\ne n\pi +\dfrac{\pi }{2},n\in \mathbb{Z}$.