Question

If \(\cos A + \sin A = \sqrt{2} \cos A\), prove that \(\cos A - \sin A = \sqrt{2} \sin A\).

Hint:

Another neat method is to square both sides of the given equation and use the identity \(\sin^2 A + \cos^2 A = 1\)!
Both methods are mathematically robust and yield full marks.

Answer 1

Step 1: Square both sides of the given equation instead of isolating sinA algebraically.
\[ (\cos A + \sin A)^2 = (\sqrt{2}\cos A)^2 \implies \cos^2A + 2\sin A\cos A + \sin^2A = 2\cos^2A \]
Step 2: Simplify using the Pythagorean identity.
Since $\sin^2A + \cos^2A = 1$:
\[ 1 + 2\sin A \cos A = 2\cos^2 A \implies 2\sin A \cos A = 2\cos^2A - 1 \]
Step 3: Build (cosA - sinA)^2 from this relation.
\[ (\cos A - \sin A)^2 = \cos^2A - 2\sin A\cos A + \sin^2A = 1 - 2\sin A\cos A = 1-(2\cos^2A-1) = 2(1-\cos^2A) = 2\sin^2A \]
Taking the square root (with the sign matching the original equation) gives:
\[ \boxed{\cos A - \sin A = \sqrt{2}\sin A} \]