Question

If $ \cos \theta + \sin \theta = \sqrt 2 \cos \theta $ . Then show, $ \cos \theta - \sin \theta = \sqrt 2 \sin \theta $

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric identities such as $ {\sin ^2}x + {\cos ^2}x = 1 $ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer:
In the given problem, we are given a trigonometric equation $ \cos \theta + \sin \theta = \sqrt 2 \cos \theta $ and we have to prove another result using the given equation.
So, we have, $ \cos \theta + \sin \theta = \sqrt 2 \cos \theta $
Squaring both sides of the equation, we get,
 $ \Rightarrow {\left( {\cos \theta + \sin \theta } \right)^2} = {\left( {\sqrt 2 \cos \theta } \right)^2} $
Now, evaluating the square of the binomial term using the algebraic identity $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $ , we get,
 $ \Rightarrow {\cos ^2}\theta + {\sin ^2}\theta + 2\cos \theta \sin \theta = 2{\cos ^2}\theta $
Shifting all the terms to right side of the equation, we get,
 $ \Rightarrow 2{\cos ^2}\theta - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta + 2\cos \theta \sin \theta } \right) = 0 $
Opening the brackets, we get,
\[ \Rightarrow 2{\cos ^2}\theta - {\cos ^2}\theta - {\sin ^2}\theta - 2\cos \theta \sin \theta = 0\]
Adding up all the like terms, we get,
\[ \Rightarrow {\cos ^2}\theta - {\sin ^2}\theta - 2\cos \theta \sin \theta = 0\]
Adding $ 2{\sin ^2}\theta $ on both sides of the equation, we get,
\[ \Rightarrow {\cos ^2}\theta - {\sin ^2}\theta - 2\cos \theta \sin \theta + 2{\sin ^2}\theta = 2{\sin ^2}\theta \]
Simplifying the equation further, we get,
\[ \Rightarrow {\cos ^2}\theta - 2\cos \theta \sin \theta + {\sin ^2}\theta = 2{\sin ^2}\theta \]
Now, using the algebraic identity $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $ , we can condense the left side of the above equation into a perfect square. So, we get,
\[ \Rightarrow {\left( {\cos \theta - \sin \theta } \right)^2} = 2{\sin ^2}\theta \]
Taking square root both sides of the equation, we get,
 \[ \Rightarrow \cos \theta - \sin \theta = \sqrt 2 \sin \theta \]
So, LHS $ = $ RHS.
Hence, $ \cos \theta - \sin \theta = \sqrt 2 \sin \theta $ if we are given $ \cos \theta + \sin \theta = \sqrt 2 \cos \theta $

Note: Given problem deals with Trigonometric functions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.