Question

If a trigonometric equation is given as $\cos x + \sin x = \sqrt 2 \cos x$ then show that $\cos x - \sin x = \sqrt 2 \sin x$ .

Answer 1

Hint: In this question, we use the basic trigonometric and algebraic identities. We have to apply some algebraic operations in the given question and then convert into the required equation. Like we use algebraic identities ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$.

Complete step-by-step solution:
Given, $\cos x + \sin x = \sqrt 2 \cos x..............\left( 1 \right)$
Now, squaring both sides in (1) equation.
$ \Rightarrow {\left( {\cos x + \sin x} \right)^2} = {\left( {\sqrt 2 \cos x} \right)^2}$
Now, we use ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$
   \Rightarrow {\cos ^2}x + {\sin ^2}x + 2 \times \cos x \times \sin x = 2{\cos ^2}x \\
   \Rightarrow 2\cos x\sin x = 2{\cos ^2}x - {\cos ^2}x - {\sin ^2}x \\
   \Rightarrow 2\cos x\sin x = {\cos ^2}x - {\sin ^2}x \\
 $
Using algebraic identity, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$ \Rightarrow 2\cos x\sin x = \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)$
From (1) equation, $\cos x + \sin x = \sqrt 2 \cos x$
$ \Rightarrow 2\cos x\sin x = \sqrt 2 \cos x\left( {\cos x - \sin x} \right)$
Now, $\cos x$ cancel from both sides.
$
   \Rightarrow \sqrt 2 \left( {\cos x - \sin x} \right) = 2\sin x \\
   \Rightarrow \cos x - \sin x = \dfrac{{2\sin x}}{{\sqrt 2 }} \\
   \Rightarrow \cos x - \sin x = \sqrt 2 \sin x \\
 $
Hence, it is proved $\cos x - \sin x = \sqrt 2 \sin x$

Note: In such types of problems we can use two different ways. First way we already mention in above and now in second way, we have to squaring both sides in given and then by using $\left( {\sin 2x = 2\sin x\cos x{\text{ and }}\cos 2x = 2{{\cos }^2}x - 1} \right)$ these identities we can get the value of x. So, after putting the value of x in question we will get the answer.