Trigonometric Equations Question 327
Question: If $ \sin \theta +\cos \theta =1 $ then the general value of $ \theta $ is
[MNR 1987; IIT 1981; Karnataka CET 2000, 03; DCE 2000]
Options:
A) $ 2n\pi $
B) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4} $
C) $ 2n\pi +\frac{\pi }{2} $
D) None of these
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Answer:
Correct Answer: B
Solution:
- $ \sin \theta +\cos \theta =1\Rightarrow \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}} $ Dividing by $ \sqrt{1^{2}+1^{2}}=\sqrt{2} $ , we get $ \sin ( \theta +\frac{\pi }{4} )=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4} $
$ \Rightarrow $ $ \theta +\frac{\pi }{4}=n\pi +{{(-1)}^{n}}\frac{\pi }{4}\Rightarrow \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4} $ .
BETA
