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

Ifcos⁡θ−sin⁡θ=2sin⁡θ\cos \theta -\sin \theta =\sqrt{2}\sin \theta , prove that cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta +\sin \theta =\sqrt{2}\cos \theta

Explanation

Solution

Hint: In such a type of question we should start our solution with the given condition. Start the solution with squaring both sides of the given condition. For this we should use the formula(a−b)2=a2+b2−2ab{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab. Meanwhile during solution, we wisely use the trigonometric identity
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 so that we can easily arrive at the required condition which we have to prove.

Complete step-by-step answer:
It is given from question
cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta -\sin \theta =\sqrt{2}\sin \theta
Here we square (involution) both side
(cos⁡θ−sin⁡θ)2=(2sin⁡θ)2{{\left( \cos \theta -\sin \theta \right)}^{2}}={{\left( \sqrt{2}\sin \theta \right)}^{2}}
As we know that
(a−b)2=a2+b2−2ab{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab
So, using the above basic algebraic formula we can expand it as
⇒cos⁡2θ+sin⁡2θ−2sin⁡θ.cos⁡θ=2sin⁡2θ\Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\sin \theta .\cos \theta =2{{\sin }^{2}}\theta
As we know that
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
So, we can write the above equation as
⇒1−2sin⁡θ.cos⁡θ=2(1−cos⁡2θ) ⇒1−2sin⁡θ.cos⁡θ=2−2cos⁡2θ \begin{aligned} & \Rightarrow 1-2\sin \theta .\cos \theta =2(1-{{\cos }^{2}}\theta ) \\\ & \Rightarrow 1-2\sin \theta .\cos \theta =2-2{{\cos }^{2}}\theta \\\ \end{aligned}

$1-2\sin \theta .\cos \theta -2+2{{\cos }^{2}}\theta =0$ This can be written as $-2\sin \theta .\cos \theta -1+2{{\cos }^{2}}\theta =0$ Now taking minus sign outside for first two terms we can write $-(2\sin \theta .\cos \theta +1)+2{{\cos }^{2}}\theta =0$ On transposing ${{\sin }^{2}}\theta $on right hand side we have $-(2\sin \theta .\cos \theta +1)=-2{{\cos }^{2}}\theta $ Multiplying both side by – sign we can write the above as $(2\sin \theta .\cos \theta +1)=2{{\cos }^{2}}\theta $ New here we can use the identity $1={{\sin }^{2}}\theta +{{\cos }^{2}}\theta $we can write further $(2\sin \theta .\cos \theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta )=2{{\cos }^{2}}\theta $ Now we can use the formula ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ So, we can write the above equation as ${{(\cos \theta +\sin \theta )}^{2}}=2{{\cos }^{2}}\theta $ Now taking square root (evolution) to the both sides viz. LSH and RHS we can write it as $\sqrt{{{(\cos \theta +\sin \theta )}^{2}}}=\sqrt{2{{\cos }^{2}}\theta }$ Taking the positive square root both side we have $(\cos \theta +\sin \theta )=\sqrt{2}\cos \theta $ Hence, we get the desired result. So, the given question is proved. Note: In the above solution we use two processes which are very important to solve such types of questions. The process is INVOLUTION that is multiplying an expression by itself, here squaring each side is INVOLUTION. Then we use the process of EVOLUTION which is defined as the root of any proposed expression is that quantity which being multiplied by itself the requisite number of times produces the given expression. Here the operation of finding the square root is Evolution.