Solving $\sum_{m=1}^{6}\csc \theta$ for $\theta$

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In summary, for $0<\theta < \frac{\pi}{2}$, the solutions to the given equation are $\frac{\pi}{12}$ and $\frac{5\pi}{12}$. These solutions can be easily verified by substituting them back into the equation.
  • #1
sbhatnagar
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For $0<\theta < \frac{\pi}{2}$, find the solution(s) of

$$\sum_{m=1}^{6}\csc \left\{ \theta +\frac{(m-1)\pi}{4}\right\}\csc \left\{ \theta +\frac{m\pi}{4}\right\}=4\sqrt{2}$$
 
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  • #2
sbhatnagar said:
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
 
  • #3
sbhatnagar said:
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.
 

FAQ: Solving $\sum_{m=1}^{6}\csc \theta$ for $\theta$

What is the definition of "Solving $\sum_{m=1}^{6}\csc \theta$ for $\theta$"?

The term "solving" in this context means finding the value of the summation expression $\sum_{m=1}^{6}\csc \theta$ for a given value of $\theta$. In other words, we are looking for the numerical result of the summation.

How do you solve $\sum_{m=1}^{6}\csc \theta$ for $\theta$?

To solve this summation, we need to use the inverse trigonometric functions. We can rewrite $\csc \theta$ as $\frac{1}{\sin \theta}$ and then use the inverse sine function to find the angle that gives us the desired value. We can do this for each term in the summation and then add up the results to get the final answer.

What are the possible values of $\theta$ in the summation $\sum_{m=1}^{6}\csc \theta$ for $\theta$?

The possible values of $\theta$ in this summation can be any real number except for values that make the denominator, $\sin \theta$, equal to zero. This includes values such as $\frac{\pi}{2}$, $\frac{3\pi}{2}$, etc. These values make the sine function undefined and the summation would not be valid.

Can you solve $\sum_{m=1}^{6}\csc \theta$ for multiple values of $\theta$?

Yes, it is possible to solve this summation for multiple values of $\theta$. However, each value of $\theta$ will give us a different result for the summation. This is because the inverse sine function will give us different angles for different values of $\csc \theta$.

Are there any special techniques or formulas that can be used to solve $\sum_{m=1}^{6}\csc \theta$ for $\theta$?

Yes, there are special formulas and identities that can be used to simplify this summation. For example, we can use the reciprocal identity for tangent, $\tan \theta = \frac{1}{\cot \theta}$, to rewrite the summation as $\sum_{m=1}^{6}\tan \theta$. This can make it easier to find the values of $\theta$ that give us the desired result. Additionally, we can use the sum and difference formulas for sine and cosine to simplify the terms in the summation.

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