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Question: If \(\cos\theta = \frac{8}{17}\) and \(\theta\) lies in the 1<sup>st</sup> quadrant, then the value ...

If cos⁡θ=817\cos\theta = \frac{8}{17} and θ\theta lies in the 1st quadrant, then the value of cos⁡(30∘+θ)+cos⁡(45∘−θ)+cos⁡(120∘−θ)\cos(30{^\circ} + \theta) + \cos(45{^\circ} - \theta) + \cos(120{^\circ} - \theta) is

A

2317(3−12+12)\frac{23}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)

B

2317(3+12+12)\frac{23}{17}\left( \frac{\sqrt{3} + 1}{2} + \frac{1}{\sqrt{2}} \right)

C

2317(3−12−12)\frac{23}{17}\left( \frac{\sqrt{3} - 1}{2} - \frac{1}{\sqrt{2}} \right)

D

2317(3+12−12)\frac{23}{17}\left( \frac{\sqrt{3} + 1}{2} - \frac{1}{\sqrt{2}} \right)

Answer

2317(3−12+12)\frac{23}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right)

Explanation

Solution

Since cos⁡θ=817\cos\theta = \frac{8}{17} and 0<θ<π2⇒sin⁡θ=1−82172=15170 < \theta < \frac{\pi}{2} \Rightarrow \sin\theta = \sqrt{1 - \frac{8^{2}}{17^{2}}} = \frac{15}{17}

The value of the given expression

=cos⁡30o.cos⁡θ−sin⁡30osin⁡θ+cos⁡45ocos⁡θ= \cos{}30^{o}.\cos\theta - \sin{}30^{o}\sin\theta + \cos{}45^{o}\cos\theta

+sin⁡45osin⁡θ+cos⁡120ocos⁡θ+sin⁡120osin⁡θ+ \sin{}45^{o}\sin\theta + \cos{}120^{o}\cos\theta + \sin{}120^{o}\sin\theta

=cos⁡θ(32+12−12)−sin⁡θ(12−12−32)= \cos\theta\left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \right) - \sin\theta\left( \frac{1}{2} - \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \right)

=817(32+12−12)+1517(32+12−12)= \frac{8}{17}\left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \right) + \frac{15}{17}\left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \right)

=2317(3−12+12)= \frac{23}{17}\left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} \right).