If cot x = frac{5}{12} for some x in (pi, fra | vedclass

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(C) Given $\cot x = \frac{5}{12}$ and $x \in (\pi, \frac{3\pi}{2})$, we have $\cos x = -\frac{5}{13}$ and $\sin x = -\frac{12}{13}$.Since $\frac{x}{2}

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$\frac{1}{\sqrt{13}}$

Vedclass · Answer

(C) Given $\cot x = \frac{5}{12}$ and $x \in (\pi, \frac{3\pi}{2})$, we have $\cos x = -\frac{5}{13}$ and $\sin x = -\frac{12}{13}$.Since $\frac{x}{2} \in (\frac{\pi}{2}, \frac{3\pi}{4})$, $\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-5/13)}{2}} = \sqrt{\frac{18/13}{2}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}$.And $\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}} = -\sqrt{\frac{1 + (-5/13)}{2}} = -\sqrt{\frac{8/13}{2}} = -\sqrt{\frac{4}{13}} = -\frac{2}{\sqrt{13}}$.The expression is $\sin 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2} + \cos 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2}$.$= (\cos 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2}) + (\sin 7x \cos \frac{13x}{2} - \cos 7x \sin \frac{13x}{2})$.$= \cos(7x - \frac{13x}{2}) + \sin(7x - \frac{13x}{2}) = \cos \frac{x}{2} + \sin \frac{x}{2}$.$= -\frac{2}{\sqrt{13}} + \frac{3}{\sqrt{13}} = \frac{1}{\sqrt{13}}$.

If $\cot x = \frac{5}{12}$ for some $x \in (\pi, \frac{3\pi}{2})$, then $\sin 7x(\cos \frac{13x}{2} + \sin \frac{13x}{2}) + \cos 7x(\cos \frac{13x}{2} - \sin \frac{13x}{2})$ is equal to

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