Question

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

Answer 1

Hint: At first, we will consider the given \[\cos \theta +\sin \theta =\sqrt{2}\cos \theta \] and take the square on both the sides and expanding identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] After that, we will subtract \[{{\cos }^{2}}\theta +2\sin \theta \cos \theta \] from both the sides and add \[{{\sin }^{2}}\theta \] both the sides to make the equation \[2{{\sin }^{2}}\theta ={{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\sin \theta \cos \theta \] after which we will use the identity \[{{a}^{2}}+{{b}^{2}}-2ab\] as \[{{\left( a-b \right)}^{2}}\] 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 \[\cos \theta +\sin \theta =\sqrt{2}\cos \theta \] and from this we have to prove that \[\cos \theta -\sin \theta =\sqrt{2}\sin \theta \]
So, at first, we will consider what is given,
\[\cos \theta +\sin \theta =\sqrt{2}\cos \theta \]
Now, at first we will square both the sides of the equation to proceed, so we get,
\[{{\left( \cos \theta +\sin \theta \right)}^{2}}={{\left( \sqrt{2}\cos \theta \right)}^{2}}\]
So, we will expand by using identity \[{{\left( a+b \right)}^{2}}\] which equals to \[{{a}^{2}}+{{b}^{2}}+2ab\] Now we will consider a as \[\cos \theta \] and b as \[\sin \theta \] so we get,
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta =2{{\cos }^{2}}\theta \]
Now, we will subtract \[{{\cos }^{2}}\theta \] from both the sides, so we get,
\[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta -{{\cos }^{2}}\theta =2{{\cos }^{2}}\theta -{{\cos }^{2}}\theta \]
Which on calculation, we get,
\[{{\sin }^{2}}\theta +2\sin \theta \cos \theta ={{\cos }^{2}}\theta \]
Now, we will subtract \[2\sin \theta \cos \theta \] from the sides, so we get,
\[\begin{align}
  & {{\sin }^{2}}\theta +2\sin \theta \cos \theta -2\sin \theta \cos \theta ={{\cos }^{2}}\theta -2\sin \theta \cos \theta \\
 & \Rightarrow \\
 & {{\sin }^{2}}\theta ={{\cos }^{2}}\theta -2\sin \theta \cos \theta \\
\end{align}\]
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.