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.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt

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.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.Expanding the expression using $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$:$\cos(30^\circ + 	heta) = \cos 30^\circ \cos 	heta - \sin 30^\circ \sin 	heta = \frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta$$\cos(45^\circ - 	heta) = \cos 45^\circ \cos 	heta + \sin 45^\circ \sin 	heta = \frac{1}{\sqrt{2}} \cos 	heta + \frac{1}{\sqrt{2}} \sin 	heta$$\cos(120^\circ - 	heta) = \cos 120^\circ \cos 	heta + \sin 120^\circ \sin 	heta = -\frac{1}{2} \cos 	heta + \frac{\sqrt{3}}{2} \sin 	heta$Summing these:$= (\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2}) \cos 	heta + (\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2}) \sin 	heta$$= (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) (\cos 	heta + \sin 	heta)$$= (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}}) (\frac{8}{17} + \frac{15}{17}) = \frac{23}{17} (\frac{\sqrt{3} - 1}{2} + \frac{1}{\sqrt{2}})$.

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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