Question

If $\sin \left( \pi \cos \theta \right)=\cos \left( \pi \sin \theta \right)$ , prove that $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{2\sqrt{2}}$.

Answer 1

For answering this question we will first simplify the equation we have $\sin \left( \pi \cos \theta \right)=\cos \left( \pi \sin \theta \right)$ using $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ . After that we will use the value we have and find the value of $\cos \left( \theta -\dfrac{\pi }{4} \right)$ using $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ .

Complete step by step answer:
From the question we have that $\sin \left( \pi \cos \theta \right)=\cos \left( \pi \sin \theta \right)$ . Since we know that the value of $\cos x$ in terms of $\sin x$ is given by $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ . By using this we will have $\cos \left( \pi \sin \theta \right)=\sin \left( \dfrac{\pi }{2}-\pi \sin \theta \right)$ .
Now by using this value in the equation we have we will have $\sin \left( \pi \cos \theta \right)=\sin \left( \dfrac{\pi }{2}-\pi \sin \theta \right)$ .
By cancelling out $\sin$ on both sides we will have $\pi \cos \theta =\dfrac{\pi }{2}-\pi \sin \theta$ .
By cancelling out $\pi$ on both sides we will have $\cos \theta =\dfrac{1}{2}-\sin \theta$ .
By transferring $\sin \theta$ from R.H.S to L.H.S we will have $\cos \theta +\sin \theta =\dfrac{1}{2}$ .
By multiplying and dividing $\sqrt{2}$ in L.H.S we will have $\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta \right)=\dfrac{1}{2}$ .
By transferring $\sqrt{2}$ from L.H.S to R.H.S we will have $\left( \dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta \right)=\dfrac{1}{2\sqrt{2}}$ .
Since we know that we need to prove that $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{2\sqrt{2}}$ we have $\cos \left( \theta -\dfrac{\pi }{4} \right)$ on the L.H.S we know that $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta$ from the expansion of $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ . Since we know that $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ .
By comparing the both equations $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta$ and $\left( \dfrac{1}{\sqrt{2}}\cos \theta +\dfrac{1}{\sqrt{2}}\sin \theta \right)=\dfrac{1}{2\sqrt{2}}$ we can say that $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{2\sqrt{2}}$.

Hence, it is proved that when $\sin \left( \pi \cos \theta \right)=\cos \left( \pi \sin \theta \right)$ the value of $\cos \left( \theta -\dfrac{\pi }{4} \right)$ is given as $\cos \left( \theta -\dfrac{\pi }{4} \right)=\dfrac{1}{2\sqrt{2}}$.

Note: We can also prove this in another way by using the inverse trigonometric functions. We can do that by applying ${{\sin }^{-1}}$ on both sides of the equation we have $\sin \left( \pi \cos \theta \right)=\cos \left( \pi \sin \theta \right)$.
By applying that we will have ${{\sin }^{-1}}\sin \left( \pi \cos \theta \right)={{\sin }^{-1}}\cos \left( \pi \sin \theta \right)\Rightarrow \pi \cos \theta =\dfrac{\pi }{2}-\pi \sin \theta$.
Since we know that ${{\sin }^{-1}}\sin x=x$ and ${{\sin }^{-1}}\cos x=\dfrac{\pi }{2}-x$ . After this the further simplifications and end up conclusion will be the same as we have done before.