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Question: If \[\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)\] , then which ...

If sin⁡(πcos⁡θ)=cos⁡(πsin⁡θ)\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right) , then which of the following is correct.
(a)cos⁡θ=322 (b)cos⁡(θ−π2)=122 (c)cos⁡(θ−π4)=122 (d)cos⁡(θ+π4)=−122  \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 }} \\\

Explanation

Solution

Hint-In this question, we use the concept of trigonometric equations. We use the general solution, cos⁡x=cos⁡y⇒x=2nπ±y,n∈I\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.

Complete step-by-step answer:
We have an equation sin⁡(πcos⁡θ)=cos⁡(πsin⁡θ)\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⁡(πcos⁡θ)=cos⁡(πsin⁡θ)\sin \left( {\pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)
As we know, sin⁡θ=cos⁡(π2−θ)\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)
⇒cos⁡(π2−πcos⁡θ)=cos⁡(πsin⁡θ)\Rightarrow \cos \left( {\dfrac{\pi }{2} - \pi \cos \theta } \right) = \cos \left( {\pi \sin \theta } \right)
Now, we apply cos⁡x=cos⁡y⇒x=2nπ±y,n∈I\cos x = \cos y \Rightarrow x = 2n\pi \pm y,n \in I
⇒π2−πcos⁡θ=2nπ±πsin⁡θ,n∈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.
⇒π2−πcos⁡θ=±πsin⁡θ ⇒12−cos⁡θ=±sin⁡θ ⇒12=cos⁡θ±sin⁡θ  \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 12\dfrac{1}{{\sqrt 2 }} on both sides of the equation.
⇒12×cos⁡θ±12×sin⁡θ=12×12\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⁡π4=sin⁡π4=12\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}
⇒cos⁡π4cos⁡θ±sin⁡π4sin⁡θ=122\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⁡π4cos⁡θ−sin⁡π4sin⁡θ=122\cos \dfrac{\pi }{4}\cos \theta - \sin \dfrac{\pi }{4}\sin \theta = \dfrac{1}{{2\sqrt 2 }}
Now we use trigonometric identity, cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B
⇒cos⁡(θ+π4)=122\Rightarrow \cos \left( {\theta + \dfrac{\pi }{4}} \right) = \dfrac{1}{{2\sqrt 2 }}
If we take positive value, cos⁡π4cos⁡θ+sin⁡π4sin⁡θ=122\cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta = \dfrac{1}{{2\sqrt 2 }}
Now we use trigonometric identity, cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B
⇒cos⁡(θ−π4)=122\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.