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Question: If \(\cos \theta +\sin \theta =\sqrt{2}\cos \theta \), show that \(\cos \theta -\sin \theta =\sqrt{2...

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

Explanation

Solution

We first take the square value of the equation cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta . We then use the identities (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab and a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to break the expression. Putting the value of cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta , we prove that cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta -\sin \theta =\sqrt{2}\sin \theta .

Complete step-by-step solution:
It is given that cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta .
We take square values on both sides of the equation.
So, we get (cos⁡θ+sin⁡θ)2=(2cos⁡θ)2{{\left( \cos \theta +\sin \theta \right)}^{2}}={{\left( \sqrt{2}\cos \theta \right)}^{2}}. We sue the identity of (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
Simplifying we get
(cos⁡θ+sin⁡θ)2=(2cos⁡θ)2 ⇒cos⁡2θ+sin⁡2θ+2cos⁡θsin⁡θ=2cos⁡2θ ⇒cos⁡2θ−sin⁡2θ=2cos⁡θsin⁡θ \begin{aligned} & {{\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 \\\ \end{aligned}
Now we use the factorisation identity of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right).
We get cos⁡2θ−sin⁡2θ=(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ){{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\left( \cos \theta +\sin \theta \right)\left( \cos \theta -\sin \theta \right).
So, we get (cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)=2cos⁡θsin⁡θ\left( \cos \theta +\sin \theta \right)\left( \cos \theta -\sin \theta \right)=2\cos \theta \sin \theta .
We place the values given of cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta and get
(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)=2cos⁡θsin⁡θ ⇒(2cos⁡θ)(cos⁡θ−sin⁡θ)=2cos⁡θsin⁡θ \begin{aligned} & \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 \\\ \end{aligned}
Dividing both sides with 2cos⁡θ\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 \\\ \end{aligned}$$ 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 $x\ne n\pi +\dfrac{\pi }{2},n\in \mathbb{Z}$.