If cos theta = frac{8}{17} and theta lies in | vedclass

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(A) Given $\cos 	heta = \frac{8}{17}$ and $	heta$ is in the $1^{st}$ quadrant,we have $\sin 	heta = \sqrt{1 - \left( \frac{8}{17} ight)^2} = \sqr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\cos 	heta = \frac{8}{17}$ and $	heta$ is in the $1^{st}$ quadrant,we have $\sin 	heta = \sqrt{1 - \left( \frac{8}{17} ight)^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.Expanding the expression $\cos (30^\circ + 	heta) + \cos (45^\circ - 	heta) + \cos (120^\circ - 	heta)$:$= (\cos 30^\circ \cos 	heta - \sin 30^\circ \sin 	heta) + (\cos 45^\circ \cos 	heta + \sin 45^\circ \sin 	heta) + (\cos 120^\circ \cos 	heta + \sin 120^\circ \sin 	heta)$$= \cos 	heta (\cos 30^\circ + \cos 45^\circ + \cos 120^\circ) + \sin 	heta (-\sin 30^\circ + \sin 45^\circ + \sin 120^\circ)$$= \cos 	heta \left( \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} ight) + \sin 	heta \left( -\frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} ight)$$= \frac{8}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} ight) + \frac{15}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} ight)$$= \left( \frac{8}{17} + \frac{15}{17} ight) \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} ight)$$= \frac{23}{17} \left( \frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}} ight)$.

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

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