Trigonometry II

[sbhatnagar]

For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$

 

[CaptainBlack]

For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$

It is fairly easy to find the solutions numerically and then to verify that they are indeed solutions, IIRC the solutions are \(\pi/12\) and \(5 \pi/12\)

CB

 

[Sudharaka]

For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$

Hi sbhatnagar,

\[\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}\]

Expanding the sum and simplification yields,

\[\frac{2(\sin\theta+\cos\theta)}{\sin\left(\theta+ \frac{\pi}{4}\right)}+\frac{\sin\theta-\cos\theta}{\cos\left(\theta+\frac{\pi}{4}\right)}=4\sqrt{2}\sin\theta\cos\theta\]

\[\Rightarrow\sin 2\theta=\frac{1}{2}\]

\[\therefore \theta=\frac{\pi}{12}\mbox{ or }\theta=\frac{5\pi}{12}\]

Kind Regards,
Sudharaka.