A) \[\cos \theta =\frac{3}{2\sqrt{2}}\]
B) \[\cos \left( \theta -\frac{\pi }{2} \right)=\frac{1}{2\sqrt{2}}\]
C) \[\cos \left( \theta -\frac{\pi }{4} \right)=\frac{1}{2\sqrt{2}}\]
D) \[\cos \left( \theta +\frac{\pi }{4} \right)=-\frac{1}{2\sqrt{2}}\]
E) \[\cos \left( \theta +\frac{\pi }{4} \right)=\frac{1}{2}\]
Correct Answer: C
Solution :
We have, \[\sin (\pi \cos \theta )=\cos (\pi \sin \theta )\] \[\Rightarrow \] \[\cos \left( \frac{\pi }{2}-\pi \cos \theta \right)=\cos (\pi \sin \theta )\] \[\Rightarrow \] \[\frac{\pi }{2}-\pi \cos \theta =2n\pi \pm \pi \sin \theta \] \[\Rightarrow \] \[\frac{1}{2}-\cos \theta =2n\pm \sin \theta \] Taking\[+\]sign \[\Rightarrow \] \[\frac{1}{2}-\cos \theta =2n+\sin \theta \] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta =\frac{1}{2\sqrt{2}}-\sqrt{2}n\] \[\Rightarrow \] \[\cos \left( \theta -\frac{\pi }{4} \right)=\frac{1}{2\sqrt{2}}\] \[(for\,n=0)\]
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