Trigonometric Identities Question 3
Question: If $ \cos \theta =\frac{8}{17} $ and $ \theta $ lies in the 1st quadrant, then the value of $ \cos (30{}^\circ +\theta )+\cos (45{}^\circ -\theta )+\cos (120{}^\circ -\theta ) $ is
Options:
A) $ \frac{23}{17}( \frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}} ) $
B) $ \frac{23}{17}( \frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}} ) $
C) $ \frac{23}{17}( \frac{\sqrt{3}-1}{2}-\frac{1}{\sqrt{2}} ) $
D) $ \frac{23}{17}( \frac{\sqrt{3}+1}{2}-\frac{1}{\sqrt{2}} ) $
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Answer:
Correct Answer: A
Solution:
Since $ \cos \theta =\frac{8}{17} $ and $ 0<\theta <\frac{\pi }{2} $
$ \Rightarrow \sin \theta =\sqrt{1-\frac{8^{2}}{17^{2}}}=\frac{15}{17} $ The value of the given expression $ =\cos 30^{o},.,\cos \theta -\sin 30^{o}\sin \theta +\cos 45^{o}\cos \theta $ $ +\sin 45^{o}\sin \theta +\cos 120^{o}\cos \theta +\sin 120^{o}\sin \theta $ $ =\cos \theta ,( \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2} )-\sin \theta ,( \frac{1}{2}-\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} ) $ $ =\frac{8}{17},( \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2} )+\frac{15}{17},( \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2} ) $ $ =\frac{23}{17},( \frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}} ) $ .
BETA
