Question

If $\sin(\pi \cos \theta) = \cos(\pi \sin \theta)$, then which of the following is correct?

• A. $\cos \theta = \dfrac{3}{2\sqrt{2}}$
• B. $\cos\!\left(\theta - \dfrac{\pi}{2}\right) = \dfrac{1}{2\sqrt{2}}$
• C. $\cos\!\left(\theta - \dfrac{\pi}{4}\right) = \dfrac{1}{2\sqrt{2}}$
• D. $\cos\!\left(\theta + \dfrac{\pi}{4}\right) = -\dfrac{1}{2\sqrt{2}}$

Hint:

$\sin A = \cos B \Rightarrow A + B = \dfrac{\pi}{2} + 2n\pi$. Use $a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}\cos(\theta - \phi)$ to simplify.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Convert $\cos$ to $\sin$ using $\cos\alpha = \sin\!\left(\dfrac{\pi}{2}-\alpha\right)$ and equate arguments.
Step 2: Detailed Explanation:
$\sin(\pi\cos\theta) = \sin\!\left(\dfrac{\pi}{2} - \pi\sin\theta\right)$.
$\Rightarrow \pi\cos\theta = \dfrac{\pi}{2} - \pi\sin\theta + 2n\pi \Rightarrow \cos\theta + \sin\theta = \dfrac{1}{2} + 2n$.
For $n=0$: $\cos\theta + \sin\theta = \dfrac{1}{2} \Rightarrow \sqrt{2}\cos\!\left(\theta-\dfrac{\pi}{4}\right) = \dfrac{1}{2} \Rightarrow \cos\!\left(\theta-\dfrac{\pi}{4}\right) = \dfrac{1}{2\sqrt{2}}$.
Step 3: Final Answer:
$\cos\!\left(\theta - \dfrac{\pi}{4}\right) = \dfrac{1}{2\sqrt{2}}$.