Question

If $\cos \theta + \sin \theta = \sqrt 2 \cos \theta $ then the value of $\cos \theta - \sin \theta $.
A. $\sqrt 2 \sin \theta $
B. $\sqrt 2 {\cos ^2}\theta $
C. $\sqrt 2 \cos \theta $
D. $\sqrt 2 {\sin ^2}\theta $

Answer 1

Hint:
We know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ for every condition and we are also given that $\cos \theta + \sin \theta = \sqrt 2 \cos \theta $ so squaring both sides we will get the value of $\cos \theta - \sin \theta $.

Complete step by step solution:
Here we are given that $\cos \theta + \sin \theta = \sqrt 2 \cos \theta $ Now first of all we must know that what is $\cos \theta {\text{ and }}\sin \theta $. They are the trigonometric functions. $\sin \theta $ is the ratio of the length of the perpendicular and the hypotenuse in the right angles triangle whereas $\cos \theta $ is the ratio of the base and the hypotenuse in the right angle triangle.
We know that the relation between the $\cos \theta {\text{ and }}\sin \theta $ is that ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Now here we are given in the question that $\cos \theta + \sin \theta = \sqrt 2 \cos \theta $
Now squaring both the sides we will get that
$\Rightarrow {(\cos \theta + \sin \theta )^2} = {(\sqrt 2 \cos \theta )^2}$
$\Rightarrow {\cos ^2}\theta + {\sin ^2}\theta + 2\sin \theta \cos \theta = 2{\cos ^2}\theta $
$\Rightarrow 1 + 2\sin \theta \cos \theta = 2{\cos ^2}\theta $
We get that
$\Rightarrow 2\sin \theta \cos \theta = 2{\cos ^2}\theta - 1$
We also know that ${(a - b)^2} = {(a + b)^2} - 4ab$
$\Rightarrow {(\cos \theta - \sin \theta )^2} = {(\cos \theta + \sin \theta )^2} - 4\sin \theta \cos \theta $
$\Rightarrow {(\cos \theta - \sin \theta )^2} = {(\sqrt 2 \cos \theta )^2} - 2(2{\cos ^2}\theta - 1)$
$\Rightarrow {(\cos \theta - \sin \theta )^2} = 2{\cos ^2}\theta - 4{\cos ^2}\theta + 2$
$\Rightarrow - 2{\cos ^2}\theta + 2$
$\Rightarrow 2(1 - {\cos ^2}\theta )$
As we know that $1 - {\cos ^2}\theta = {\sin ^2}\theta $
So we get that
$\Rightarrow {(\cos \theta - \sin \theta )^2} = 2{\sin ^2}\theta $
Taking the square root we get that
$\Rightarrow \cos \theta - \sin \theta = \sqrt 2 \sin \theta $
As we know that $2\sin \theta \cos \theta = \sin 2\theta $
And $\Rightarrow 2{\cos ^2}\theta - 1 = \cos 2\theta $
Therefore we must say that
$\sin 2\theta = \cos 2\theta $
Both are equal at $45^\circ $
So we can say that
$
\Rightarrow 2\theta = 45^\circ \\
\Rightarrow \theta = \dfrac{{45^\circ }}{2} \\
 $
So $(\cos \theta - \sin \theta ) = \sqrt 2 \sin \theta $

So option A is the correct option.

Note:
We can do it by this method also as we are given that $\cos \theta + \sin \theta = \sqrt 2 \cos \theta $
So we can write that $\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta = \cos \theta $
As we know that $\cos 45 = \sin 45 = \dfrac{1}{{\sqrt 2 }}$
So we can say that $\cos 45.\cos \theta + \sin 45.\sin \theta = \cos \theta $
And we know that $\cos A\cos B + \sin A\sin B = \cos (A - B)$
$\cos (45 - \theta ) = \cos \theta $
$
\Rightarrow 45 - \theta = \theta \\ \Rightarrow
  2\theta = 45 \\ \Rightarrow
  \theta = \dfrac{{45}}{2} \\
 $
Hence we can say that $(\cos \theta - \sin \theta ) = \sqrt 2 \sin \theta $
Hence we get the same answer.