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Question: If we have a trigonometric equation as \[\cos \theta +\sin \theta =\sqrt{2}\cos \theta \] , then sho...

If we have a trigonometric equation as cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta , then show that cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta -\sin \theta =\sqrt{2}\sin \theta

Explanation

Solution

At first, we will consider the given cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta and take the square on both the sides and expanding identity (a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} After that, we will subtract cos⁡2θ+2sin⁡θcos⁡θ{{\cos }^{2}}\theta +2\sin \theta \cos \theta from both the sides and add sin⁡2θ{{\sin }^{2}}\theta both the sides to make the equation 2sin⁡2θ=cos⁡2θ+sin⁡2θ−2sin⁡θcos⁡θ2{{\sin }^{2}}\theta ={{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\sin \theta \cos \theta after which we will use the identity a2+b2−2ab{{a}^{2}}+{{b}^{2}}-2ab as (a−b)2{{\left( a-b \right)}^{2}} and take square root on both the sides to get the answer.

Complete step-by-step answer:
In the question, an equation is given as cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta and from this we have to prove that cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta -\sin \theta =\sqrt{2}\sin \theta
So, at first, we will consider what is given,
cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta
Now, at first we will square both the sides of the equation to proceed, 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}}
So, we will expand by using identity (a+b)2{{\left( a+b \right)}^{2}} which equals to a2+b2+2ab{{a}^{2}}+{{b}^{2}}+2ab Now we will consider a as cos⁡θ\cos \theta and b as sin⁡θ\sin \theta so we get,
cos⁡2θ+sin⁡2θ+2sin⁡θcos⁡θ=2cos⁡2θ{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta =2{{\cos }^{2}}\theta
Now, we will subtract cos⁡2θ{{\cos }^{2}}\theta from both the sides, so we get,
cos⁡2θ+sin⁡2θ+2sin⁡θcos⁡θ−cos⁡2θ=2cos⁡2θ−cos⁡2θ{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta -{{\cos }^{2}}\theta =2{{\cos }^{2}}\theta -{{\cos }^{2}}\theta
Which on calculation, we get,
sin⁡2θ+2sin⁡θcos⁡θ=cos⁡2θ{{\sin }^{2}}\theta +2\sin \theta \cos \theta ={{\cos }^{2}}\theta
Now, we will subtract 2sin⁡θcos⁡θ2\sin \theta \cos \theta from the sides, so we get,

& {{\sin }^{2}}\theta +2\sin \theta \cos \theta -2\sin \theta \cos \theta ={{\cos }^{2}}\theta -2\sin \theta \cos \theta \\\ & \Rightarrow \\\ & {{\sin }^{2}}\theta ={{\cos }^{2}}\theta -2\sin \theta \cos \theta \\\ \end{aligned}$$ Now, we will add $${{\sin }^{2}}\theta $$ to both the sides, so we get, $$2{{\sin }^{2}}\theta ={{\cos }^{2}}\theta -2\sin \theta \cos \theta +{{\sin }^{2}}\theta $$ We know an identity which is $${{\left( a-b \right)}^{2}}$$ which equals to $${{a}^{2}}+{{b}^{2}}-2ab$$ and we will use it as $${{a}^{2}}+{{b}^{2}}-2ab$$ equals to $${{\left( a-b \right)}^{2}}$$ where a is $$\cos \theta $$ and b is $$\sin \theta $$ so we get, $$2{{\sin }^{2}}\theta ={{\left( \cos \theta -\sin \theta \right)}^{2}}$$ Which can also be written as, $${{\left( \cos \theta -\sin \theta \right)}^{2}}=2{{\sin }^{2}}\theta $$ Now, on taking square root on both the sides, we get, $$\cos \theta -\sin \theta =\sqrt{2}\sin \theta $$ So, hence it is proved. **Note:** We can do the same question by another method. We can consider the following identity, $${{\left( a-b \right)}^{2}}+{{\left( a+b \right)}^{2}}=4ab$$ where a is $$\cos \theta $$ and b is $$\sin \theta $$ and we are given value of $$\cos \theta +\sin \theta \text{ as }\sqrt{2}\cos \theta $$ we will find value of $$\cos \theta -\sin \theta $$ to get the answer.