Question

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

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

Answer 1

The correct option is A

$(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\frac{23}{17}$

$cos\theta =\frac{8}{17}\Rightarrow sin\theta =\frac{\sqrt{(17{)}^2-8^2}}{17}=\frac{15}{17}$ Now the given expression is equal to

$cos{30}^{\circ }cos\theta -sin{30}^{\circ }sin\theta +cos{45}^{\circ }cos\theta +sin{45}^{\circ }sin\theta +cos{120}^{\circ }cos\theta +sin{120}^{\circ }sin\theta =cos\theta (cos{30}^{\circ }+cos{45}^{\circ }+cos{120}^{\circ })-sin\theta (sin{30}^{\circ }-sin{45}^{\circ }-sin{120}^{\circ })$

$=\frac{8}{17}(\frac{\sqrt{3}}{7}+\frac{1}{\sqrt{2}})-\frac{15}{17}(\frac{1}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2})=(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}})\frac{23}{17}$