Question
If $cotx = \frac{5}{12}$ for some $x \in \left( \pi,\frac{3\pi}{2} \right)$, then
$$sin7x\left( cos\frac{13x}{2} + sin\frac{13x}{2} \right) + $$$cos7x\left( cos\frac{13x}{2} - sin\frac{13x}{2} \right)$ is equal to
Options
Solution
$cotx = \frac{5}{12} \Rightarrow cosx = \frac{- 5}{13} = 2\cos^{2}\frac{x}{2} - 1$
$cos\left( \frac{x}{2} \right) = - \frac{2}{\sqrt{13}}$ or $\frac{2}{\sqrt{13}}($ rejected $)$
$${\left\{ \because\frac{x}{2} \in \left( \frac{\pi}{2},\frac{3\pi}{4} \right) \right\} }{\left( sin7x\frac{sin13x}{2} + cos7x\frac{cos13x}{2} \right) + \left( sin7x\frac{cos13x}{2} - cos7x\frac{sin13x}{2} \right)cos\left( 7x - \frac{13x}{2} \right) + sin\left( 7x - \frac{13x}{2} \right) }{cos\frac{x}{2} + sin\left( \frac{x}{2} \right) }{\frac{3}{\sqrt{13}} - \frac{2}{\sqrt{13}} = \frac{1}{\sqrt{13}}}$$
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