Let ( )/(2)

Question

Let $\frac{\pi }{2}$

Accepted Answer

$\sin (\frac{15\theta }{2})(\cos 8\theta +\sin 8\theta )+\cos (\frac{15\theta }{2})(\cos 8\theta -\sin 8\theta )$

= $\sin (\frac{15\theta }{2})\cos 8\theta +\sin (\frac{15\theta }{2})\sin 8\theta +\cos (\frac{15\theta }{2})\cos 8\theta -\cos (\frac{15\theta }{2})\sin 8\theta $

= $[\cos \frac{15\theta }{2}\cos 8\theta +\sin \frac{15\theta }{2}\sin 8\theta ]+[\sin \frac{15\theta }{2}\cos 8\theta -\cos \frac{15\theta }{2}\sin 8\theta ]$

= $\cos (8\theta -\frac{15\theta }{2})+\sin (\frac{15\theta }{2}-8\theta )$

= $\cos (\frac{\theta }{2})+\sin (-\frac{\theta }{2})=\cos \frac{\theta }{2}-\sin \frac{\theta }{2}$

$\cot \theta =-\frac{1}{2\sqrt{2}},\frac{\pi }{2}$\cos \theta =-\frac{1}{3}$

$$$\begin{gathered} \cos \theta =1-2{\sin }^{2}\frac{\theta }{2}\implies -\frac{1}{3}=1-2{\sin }^{2}\frac{\theta }{2}\implies \sin \frac{\theta }{2}=\sqrt{\frac{2}{3}} \ \cos \theta =2{\cos }^{2}\frac{\theta }{2}-1\implies -\frac{1}{3}=2{\cos }^{2}\frac{\theta }{2}-1\implies \cos \frac{\theta }{2}=\sqrt{\frac{1}{3}} \end{gathered}$$$

$\therefore $ $\cos \frac{\theta }{2}-\sin \frac{\theta }{2}$

$$$\begin{gathered} =\sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}} \ =\frac{1-\sqrt{2}}{\sqrt{3}} \end{gathered}$$

Solutions · 1