A) \[\theta ={{30}^{o}}\]
B) \[\theta ={{45}^{o}}\]
C) \[\theta ={{60}^{o}}\]
D) \[\theta ={{90}^{o}}\]
Correct Answer: B
Solution :
Let \[f(x)=\sin \theta +\cos \theta =\sqrt{2}\sin \left( \theta +\frac{\pi }{4} \right)\] But\[-1\le \sin \left( \theta +\frac{\pi }{2} \right)\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\sin \left( \theta +\frac{\pi }{4} \right)\le \sqrt{2}\]. Hence the maximum value of \[(\sin \theta +\cos \theta )\] i.e., of \[\sqrt{2}\sin \left( \theta +\frac{\pi }{4} \right)=\sqrt{2}\]. \[\therefore \]\[\sin \left( \theta +\frac{\pi }{4} \right)=1\Rightarrow \sin \left( \theta +\frac{\pi }{4} \right)=\sin \frac{\pi }{2}\] Þ \[\theta +\frac{\pi }{4}=\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}={{45}^{o}}\].
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