Question

If $\cos \theta =\frac{8}{17}$ and $\theta$ lies in the $1$st quadrant, then the value of $\cos (30°+\theta )+\cos (45°-\theta )+\cos (120°-\theta )$ is

• A. $\left(\frac{23}{17}\right)\left[\frac{(\sqrt{3}–1)}{2}+\frac{1}{\sqrt{2}}\right]$
• B. $\left(\frac{23}{17}\right)\left[\frac{(\sqrt{3}+1)}{2}+\frac{1}{\sqrt{2}}\right]$
• C. $\left(\frac{23}{17}\right)\left[\frac{(\sqrt{3}–1)}{2}-\frac{1}{\sqrt{2}}\right]$
• D. $\left(\frac{23}{17}\right)\left[\frac{(\sqrt{3}+1)}{2}-\frac{1}{\sqrt{2}}\right]$

Answer 1

The correct option is A

$\left(\frac{23}{17}\right)\left[\frac{(\sqrt{3}–1)}{2}+\frac{1}{\sqrt{2}}\right]$

Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{30}\mathbf{°}\mathbf{}\mathbf{+}\mathbf{}\mathit{\theta }\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{45}\mathbf{°}\mathbf{}\mathbf{-}\mathbf{}\mathit{\theta }\mathbf{)}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{120}\mathbf{°}\mathbf{}\mathbf{-}\mathbf{}\mathit{\theta }\mathbf{)}$:

Given, $\cos \theta =\frac{8}{17}$ and $\theta$ lies in the $1st$ quadrant

$\Rightarrow$ $\sin \theta =\frac{15}{17}$

$\therefore$$\cos (30°+\theta )+\cos (45°-\theta )+\cos (120°-\theta )$

$$\begin{gathered} =\cos 30°\cos \theta –\sin 30°\sin \theta +\cos 45°\cos \theta +\sin 45°\sin \theta +\cos 120°\cos \theta +\sin 120°\sin \theta \\ =\left(\frac{\sqrt{3}}{2}\right)\cos \theta –\left(\frac{1}{2}\right)\sin \theta +\left(\frac{1}{\sqrt{2}}\right)\cos \theta +\left(\frac{1}{\sqrt{2}}\right)\sin \theta +\left(-\frac{1}{2}\right)\cos \theta +\left(\frac{\sqrt{3}}{2}\right)\sin \theta \\ =\left[\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)–\left(\frac{1}{2}\right)\right]\cos \theta +\left[\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\right]\sin \theta \\ =\left[\frac{(\sqrt{6}+2–\sqrt{2})}{2\sqrt{2}}\right]\left(\frac{8}{17}\right)+\left[(2–\sqrt{2}+\sqrt{6})/2\sqrt{2}\right]\left(\frac{15}{17}\right) \\ =\left[\frac{1}{17\left(2\sqrt{2}\right)}\right]\left(8\sqrt{6}+16–8\sqrt{2}+30–15\sqrt{2}+15\sqrt{6}\right) \\ =\left[\frac{1}{17\left(2\sqrt{2}\right)}\right](23\sqrt{6}–23\sqrt{2}+46) \\ =\left[\frac{23\sqrt{6}}{17\left(2\sqrt{2}\right)}\right]–\left[\frac{23\sqrt{2}}{17\left(2\sqrt{2}\right)}\right]+\left[\frac{46}{17\left(2\sqrt{2}\right)}\right] \\ =\left[\frac{23\sqrt{3}}{17\left(2\right)}\right]–\left[\frac{23}{17\left(2\right)}\right]+\left[\frac{46}{17\left(2\sqrt{2}\right)}\right] \\ =\mathbf{\left(}\frac{\mathbf{23}}{\mathbf{17}}\mathbf{\right)}\mathbf{}\mathbf{[}\frac{\mathbf{(}\sqrt{\mathbf{3}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{]} \end{gathered}$$

Hence, Option ‘A’ is Correct.