Trigonometric Identities Question 269
Question: The value $ \cos 105{}^\circ +\sin 105{}^\circ $ is
[MNR 1975]
Options:
A) $ \frac{1}{2} $
1
C) $ \sqrt{2} $
D) $ \frac{1}{\sqrt{2}} $
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Answer:
Correct Answer: D
Solution:
$ \cos 105{}^\circ +\sin 10+\cos 105{}^\circ +\sin 105{}^\circ =\cos (90{}^\circ +15{}^\circ )+\sin (90{}^\circ +15{}^\circ ) $ = $ -\sin 15{}^\circ +\cos 15{}^\circ =\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}} $ .
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