Solveeit Logo
Login

Question

Question: If \( \cos \theta + \sin \theta = \sqrt 2 \cos \theta \) . Then show, \( \cos \theta - \sin \theta =...

If cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta + \sin \theta = \sqrt 2 \cos \theta . Then show, cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta - \sin \theta = \sqrt 2 \sin \theta

Explanation

Solution

Hint : The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric identities such as sin⁡2x+cos⁡2x=1{\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⁡θ+sin⁡θ=2cos⁡θ\cos \theta + \sin \theta = \sqrt 2 \cos \theta and we have to prove another result using the given equation.
So, we have, cos⁡θ+sin⁡θ=2cos⁡θ\cos \theta + \sin \theta = \sqrt 2 \cos \theta
Squaring both sides of the equation, we get,
⇒(cos⁡θ+sin⁡θ)2=(2cos⁡θ)2\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 (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} , we get,
⇒cos⁡2θ+sin⁡2θ+2cos⁡θsin⁡θ=2cos⁡2θ\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,
⇒2cos⁡2θ−(cos⁡2θ+sin⁡2θ+2cos⁡θsin⁡θ)=0\Rightarrow 2{\cos ^2}\theta - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta + 2\cos \theta \sin \theta } \right) = 0
Opening the brackets, we get,
⇒2cos⁡2θ−cos⁡2θ−sin⁡2θ−2cos⁡θsin⁡θ=0\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,
⇒cos⁡2θ−sin⁡2θ−2cos⁡θsin⁡θ=0\Rightarrow {\cos ^2}\theta - {\sin ^2}\theta - 2\cos \theta \sin \theta = 0
Adding 2sin⁡2θ2{\sin ^2}\theta on both sides of the equation, we get,
⇒cos⁡2θ−sin⁡2θ−2cos⁡θsin⁡θ+2sin⁡2θ=2sin⁡2θ\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,
⇒cos⁡2θ−2cos⁡θsin⁡θ+sin⁡2θ=2sin⁡2θ\Rightarrow {\cos ^2}\theta - 2\cos \theta \sin \theta + {\sin ^2}\theta = 2{\sin ^2}\theta
Now, using the algebraic identity (a−b)2=a2−2ab+b2{\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,
⇒(cos⁡θ−sin⁡θ)2=2sin⁡2θ\Rightarrow {\left( {\cos \theta - \sin \theta } \right)^2} = 2{\sin ^2}\theta
Taking square root both sides of the equation, we get,
⇒cos⁡θ−sin⁡θ=2sin⁡θ\Rightarrow \cos \theta - \sin \theta = \sqrt 2 \sin \theta
So, LHS == RHS.
Hence, cos⁡θ−sin⁡θ=2sin⁡θ\cos \theta - \sin \theta = \sqrt 2 \sin \theta if we are given cos⁡θ+sin⁡θ=2cos⁡θ\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.