Trigonometry II

sbhatnagar

Active member
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

Well-known member
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

Well-known member
MHB Math Helper
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.