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

Question

If $\displaystyle 0 < \theta < \frac{\pi}{2}$ and $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\}\cdot \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$. Then value of $\theta=.$

$\bf{My\; Try::}$ Here the question is given in the form of $\displaystyle \frac{1}{\sin A \cdot \sin B}$. Then we can write it as

$$\displaystyle \Rightarrow \frac{1}{\sin A \cdot \sin B} = \frac{1}{\sin (B-A)}\cdot \left[\frac{\sin (B-A)}{\sin A\cdot \sin B}\right] $$

$$\displaystyle \Rightarrow \frac{1}{\sin(B-A)}\cdot \left[\frac{\cos A\cdot \sin B-\sin A\cdot \cos B}{\sin A \cdot \sin B}\right]$$

$$\displaystyle = \frac{1}{\sin (B-A)}\cdot \left[\cot A-\cot B\right]$$

Now Let $\displaystyle \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}=A$ and $\displaystyle \left\{\theta+m\cdot \frac{\pi}{4}\right\}=B.$

So We Get $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin A \cdot \sin B}=\sum_{m=1}^{6}\sqrt{2}\left[\cot A-\cot B\right] = 4\sqrt{2}.$

So we get $$\displaystyle \sum_{m=1}^{6}\left[\cot \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}-\cot \left\{\theta+m\cdot \frac{\pi}{4}\right\}\right]=4.$$

Now after expanding we get $$\displaystyle \cot \theta - \cot \left(\theta+\frac{6\pi}{4}\right) = 4.\Rightarrow \cot \theta - \cot \left(\theta+\frac{2\pi}{3}\right)=4$$

Again after simplifying we get $$\displaystyle \cot \theta - \left(\frac{\cot \theta-\sqrt{3}}{1-\sqrt{3}\cot \theta}\right)=4\Rightarrow \frac{\sqrt{3}-\sqrt{3}\cot^2 \theta}{1-\sqrt{3}\cot \theta} = 4.$$

So We get $$\displaystyle \sqrt{3}-\sqrt{3}\cot^2 \theta = 4-4\sqrt{3}\cot \theta.\Rightarrow \sqrt{3}\cot^2 \theta - 4\sqrt{3}\cot \theta+(4-\sqrt{3})=0$$.

Now Let $\cot \theta = y\;,$ Then equation is $\sqrt{3}y^2-4\sqrt{3}y+(4-\sqrt{3}) = 0.$

We Get $$\displaystyle y = \cot \theta = \frac{2\pm \sqrt{15-4\sqrt{3}}}{3}$$

But I did not understand how can i get value of $\theta.$

Help me, Thanks

Answer 1

HINT:

Can you try tis way?

$$\cot\theta-\cot\left(\frac{2\pi}3+\theta\right)=\frac{\sin\dfrac{2\pi}3}{\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}=\frac{\sqrt3}{2\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}$$

$$=\frac{\sqrt3}{\cos\dfrac{2\pi}3+\cos\left(\dfrac{2\pi}3+2\theta\right)}$$

Answer 2

Given that,

$\sf \: \displaystyle\sum^{6}_{m = 1} \frac{1}{ \sin \{ \theta + (m - 1) \frac{\pi}{4} \} \sin \{ \theta + m \frac{\pi}4 \}} = 4 \sqrt{2}$

can be rewritten as

$\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \frac{1}{ \sin \{ \theta + m\frac{\pi}{4} - \frac{\pi}{4} \} \sin \{ \theta + m \frac{\pi}4 \}} = 4 \sqrt{2} \\ \end{gathered}$

Let assume that

$\begin{gathered}\sf \: \theta + m\frac{\pi}{4} = x \\ \end{gathered}$

So, above expression can be rewritten as

$\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \frac{1}{ \sin \{ x - \frac{\pi}{4} \} \sin \{ x\}} = 4 \sqrt{2} \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \frac{\sin \frac{\pi}{4}}{ \sin \{ x - \frac{\pi}{4} \} \sin \{ x\}} = 4 \sqrt{2} \times \sin \frac{\pi}{4}\\ \end{gathered}$

$\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \frac{\sin \{x-(x-\frac{\pi}{4})\}}{ \sin \{ x - \frac{\pi}{4} \} \sin \{ x\}} = 4 \sqrt{2} \times \frac{1}{\sqrt{2}}\\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\bf \: \dfrac{sin(x - y)}{sinx \: siny} = coty - cotx \: } \\ \end{gathered}$

So, using this result, we get

$\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \left(cot\left[ x - \dfrac{\pi}{4}\right] - cotx\right) = 4 \\ \end{gathered}$

On substituting the value of x, we get

\begin{gathered}\sf\: \displaystyle\sum^{6}_{m = 1} \left(cot\left[\theta + m\dfrac{\pi}{4}- \dfrac{\pi}{4}\right] - cot\left[\theta + \dfrac{\pi}{4} \right]\right) = 4 \\ \end{gathered}

On substituting the value of m = 1, 2, ..., 6, we get

\begin{gathered}{\sf \: cot\theta - cot\left[\theta + \dfrac{\pi}{4} \right] + cot\left[\theta + \dfrac{\pi}{4} \right]- cot\left[\theta + \dfrac{2\pi}{4} \right] + ... + cot\left[\theta + \dfrac{5\pi}{4} \right] - cot\left[\theta + \dfrac{6\pi}{4} \right] = 4} \\ \end{gathered}

\begin{gathered}{\sf \: cot\theta - cot\left[\theta + \dfrac{3\pi}{2} \right] =4}\\ \end{gathered}

$\cotθ+\tanθ=4$

​ $\dfrac{1}{\tan \theta}+{\tan \theta}=4$

$\tan²θ+1=4\tanθ$

​ $\tan²θ−4\tanθ=−1$ ​

$\tan²θ−4\tanθ+4=4−1$

$(\tan \theta-2)²=3$

$(\tan \theta-2)=\pm\sqrt{3}$

\begin{gathered}\sf \: {tan\theta } = 2 +\sqrt{3} \: \: or \: \: {tan\theta } = 2 - \sqrt{3} \\ \end{gathered}

\begin{gathered}\sf \: {tan\theta } = \tan\frac{5\pi}{12} \: \: or \: \: \tan\frac{\pi}{12}\\ \end{gathered}

\begin{gathered}\sf \: {\theta } = \frac{5\pi}{12} \: \: or \: \: \frac{\pi}{12}\\ \end{gathered}