Trigonometric Identities Question 360
Question: The value of $ \sin \theta +\cos \theta $ will be greatest when
[MNR 1977, 1983; RPET 1995]
Options:
A) $ \theta =30^{o} $
B) $ \theta =45^{o} $
C) $ \theta =60^{o} $
D) $ \theta =90^{o} $
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Answer:
Correct Answer: B
Solution:
Let $ f(x)=\sin \theta +\cos \theta =\sqrt{2}\sin ( \theta +\frac{\pi }{4} ) $ But $ -1\le \sin ( \theta +\frac{\pi }{2} )\le 1\Rightarrow -\sqrt{2}\le \sqrt{2}\sin ( \theta +\frac{\pi }{4} )\le \sqrt{2} $ . Hence the maximum value of $ (\sin \theta +\cos \theta ) $ i.e., of $ \sqrt{2}\sin ( \theta +\frac{\pi }{4} )=\sqrt{2} $ .
$ \therefore $ $ \sin ( \theta +\frac{\pi }{4} )=1\Rightarrow \sin ( \theta +\frac{\pi }{4} )=\sin \frac{\pi }{2} $
Þ $ \theta +\frac{\pi }{4}=\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}=45^{o} $ .
BETA
