Question

If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, then $\cos \theta - \sin \theta =$

• A. $\sqrt{2} \sin \theta$
• B. $-\sqrt{2} \sin \theta$
• C. $\sqrt{2} \cos \theta$
• D. $-\sqrt{2} \cos \theta$

Hint:

There is a famous identity: if $a \cos\theta + b \sin\theta = c$, then $b \cos\theta - a \sin\theta = \pm \sqrt{a^2+b^2-c^2}$. Here, $1^2 + 1^2 - (\sqrt{2}\cos\theta)^2 = 2 - 2\cos^2\theta = 2\sin^2\theta$, giving $\sqrt{2}\sin\theta$.

Answer 1

The Correct Option is A

Step 1: Start from the given relation.
We have $\sin\theta + \cos\theta = \sqrt2\cos\theta$. We want a value for $\cos\theta - \sin\theta$.

Step 2: Isolate $\sin\theta$.
Move $\cos\theta$ across: \[ \sin\theta = (\sqrt2 - 1)\cos\theta \]

Step 3: A clever multiplier.
Multiply both sides by $(\sqrt2 + 1)$, because $(\sqrt2-1)(\sqrt2+1) = 1$ tidies the right side.

Step 4: Expand.
\[ (\sqrt2 + 1)\sin\theta = (2 - 1)\cos\theta = \cos\theta \]

Step 5: Open the left side.
\[ \sqrt2\sin\theta + \sin\theta = \cos\theta \]

Step 6: Rearrange to the target.
Move $\sin\theta$ over: \[ \cos\theta - \sin\theta = \sqrt2\sin\theta \] \[ \boxed{ \cos\theta - \sin\theta = \sqrt2\,\sin\theta } \]