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# If $\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)$ , then which of the following is correct.$\left( a \right)\cos \theta = \dfrac{3}{{2\sqrt 2 }} \\ \left( b \right)\cos \left( {\theta - \dfrac{\pi }{2}} \right) = \dfrac{1}{{2\sqrt 2 }} \\ \left( c \right)\cos \left( {\theta - \dfrac{\pi }{4}} \right) = \dfrac{1}{{2\sqrt 2 }} \\ \left( d \right)\cos \left( {\theta + \dfrac{\pi }{4}} \right) = - \dfrac{1}{{2\sqrt 2 }} \\$

Hint-In this question, we use the concept of trigonometric equations. We use the general solution, $\cos x = \cos y \Rightarrow x = 2n\pi \pm y,n \in I$ . First we try to write a question in terms of the general equation then apply the formula of the trigonometric equation.

We have an equation $\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)$ and we have to write equation in term of general trigonometric equation.
$\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)$
As we know, $\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)$
$\Rightarrow \cos \left( {\dfrac{\pi }{2} - \pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)$
Now, we apply $\cos x = \cos y \Rightarrow x = 2n\pi \pm y,n \in I$
$\Rightarrow \dfrac{\pi }{2} - \pi \cos \theta = 2n\pi \pm \pi \sin \theta ,n \in I$
This relation holds for any integer but now we use n=0 for easy calculation.
$\Rightarrow \dfrac{\pi }{2} - \pi \cos \theta = \pm \pi \sin \theta \\ \Rightarrow \dfrac{1}{2} - \cos \theta = \pm \sin \theta \\ \Rightarrow \dfrac{1}{2} = \cos \theta \pm \sin \theta \\$
Now, we multiply by $\dfrac{1}{{\sqrt 2 }}$ on both sides of the equation.
$\Rightarrow \dfrac{1}{{\sqrt 2 }} \times \cos \theta \pm \dfrac{1}{{\sqrt 2 }} \times \sin \theta = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
As we know, $\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \cos \dfrac{\pi }{4}\cos \theta \pm \sin \dfrac{\pi }{4}\sin \theta = \dfrac{1}{{2\sqrt 2 }}$
If we take negative value, $\cos \dfrac{\pi }{4}\cos \theta - \sin \dfrac{\pi }{4}\sin \theta = \dfrac{1}{{2\sqrt 2 }}$
Now we use trigonometric identity, $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
$\Rightarrow \cos \left( {\theta + \dfrac{\pi }{4}} \right) = \dfrac{1}{{2\sqrt 2 }}$
If we take positive value, $\cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta = \dfrac{1}{{2\sqrt 2 }}$
Now we use trigonometric identity, $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
$\Rightarrow \cos \left( {\theta - \dfrac{\pi }{4}} \right) = \dfrac{1}{{2\sqrt 2 }}$
So, the correct option is (c).
Note-In such types of problems we use some important points to solve questions in an easy way. First we use trigonometric general equations and equate angles for integer n=0 for easy calculation. Then we use trigonometric identities to solve the equation. So, we will get the required answer.