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Question: If \(\cos A + \sin A = \sqrt 2 \cos A\), then \(\cos A - \sin A\) is: A.\[\sqrt 2 \sin A\] B.\[2...

If cos⁡A+sin⁡A=2cos⁡A\cos A + \sin A = \sqrt 2 \cos A, then cos⁡A−sin⁡A\cos A - \sin A is:
A.2sin⁡A\sqrt 2 \sin A
B.2sin⁡A2\sin A
C.−2sin⁡A - \sqrt 2 \sin A
D.−2sin⁡A - 2\sin A

Explanation

Solution

We begin by squaring both sides of the given equation. Then, use the identity cos⁡2a+sin⁡2a=1{\cos ^2}a + {\sin ^2}a = 1 and substitute the required values. Next, rearrange the equation such that we get an expression equal to (cos⁡A−sin⁡A)2{\left( {\cos A - \sin A} \right)^2}. Next, take the square root on both the sides to get the required answer.

Complete step-by-step answer:
We are given that, cos⁡A+sin⁡A=2cos⁡A\cos A + \sin A = \sqrt 2 \cos A
We will square both sides of the given equation
(cos⁡A+sin⁡A)2=(2cos⁡A)2{\left( {\cos A + \sin A} \right)^2} = {\left( {\sqrt 2 \cos A} \right)^2}
Then, simplify the expression using the formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
Hence, we have
cos⁡2A+sin⁡2A+2sin⁡Acos⁡A=2cos⁡2A{\cos ^2}A + {\sin ^2}A + 2\sin A\cos A = 2{\cos ^2}A
Next, we know that cos⁡2a+sin⁡2a=1{\cos ^2}a + {\sin ^2}a = 1 which implies cos⁡2A=1−sin⁡2A{\cos ^2}A = 1 - {\sin ^2}A and sin⁡2A=1−cos⁡2A{\sin ^2}A = 1 - {\cos ^2}A
We will substitute the values of cos⁡2A{\cos ^2}A and sin⁡2A{\sin ^2}A
Therefore, we have
cos⁡2A+sin⁡2A+2sin⁡Acos⁡A=2cos⁡2A 1−sin⁡2A+1−cos⁡2A+2sin⁡Acos⁡A=2cos⁡2A  {\cos ^2}A + {\sin ^2}A + 2\sin A\cos A = 2{\cos ^2}A \\\ 1 - {\sin ^2}A + 1 - {\cos ^2}A + 2\sin A\cos A = 2{\cos ^2}A \\\
We wish to make the formula (cos⁡A−sin⁡A)2{\left( {\cos A - \sin A} \right)^2}, therefore rearrange the terms such that we get an expression equal to (cos⁡A−sin⁡A)2{\left( {\cos A - \sin A} \right)^2}
2−2cos⁡2A=cos⁡2A+sin⁡2A−2sin⁡Acos⁡A 2(1−cos⁡2A)=(cos⁡A−sin⁡A)2  2 - 2{\cos ^2}A = {\cos ^2}A + {\sin ^2}A - 2\sin A\cos A \\\ 2\left( {1 - {{\cos }^2}A} \right) = {\left( {\cos A - \sin A} \right)^2} \\\
Also, sin⁡2A=1−cos⁡2A{\sin ^2}A = 1 - {\cos ^2}A, substitute the value and take square root of the equation, then we will get,
2sin⁡2A=(cos⁡A−sin⁡A)2 cos⁡A−sin⁡A=2sin⁡A  2{\sin ^2}A = {\left( {\cos A - \sin A} \right)^2} \\\ \cos A - \sin A = \sqrt 2 \sin A \\\
Hence, option A is the correct answer.

Note For these types of questions, students must remember the formulas and identities of trigonometry. The approach used in this question is that we find the value of (cos⁡A+sin⁡A)2{\left( {\cos A + \sin A} \right)^2} by squaring the given condition and then use various trigonometry identities to find the value of (cos⁡A−sin⁡A)2{\left( {\cos A - \sin A} \right)^2} followed by taking square root of both sides.