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I just stumbled accross this question and I am unable to move ahead in the problem.

What is the value of the following summation? $$\sum_{r=1}^{88}{(-1)^{r+1}\frac{1}{\sin^2(r+1^\circ)-\sin^2(1^\circ)}}$$

MY APPROACH:

Step 1: Expand.

This is what I usually do when I come accross a summation till a specific number.

$$\begin{align} \sum_{r=1}^{88}{(-1)^{r+1} \frac{1}{\sin^2(r+1^\circ)- \sin^2(1^\circ)}} &= \frac{1}{\sin^2(2^\circ)-\sin^2(1^\circ)}-\frac{1}{\sin^2(3^\circ)-\sin^2(1^\circ)} \\ &\quad+\frac{1}{\sin^2(4^\circ)-\sin^2(1^\circ)}\cdots-\frac{1}{\sin^2(89^\circ)-\sin^2(1^\circ)} \\ &= \frac{1}{[\sin(2^\circ)+\sin(1^\circ)][\sin(2^\circ)-\sin(1^\circ)]} \\ &\quad-\frac{1}{[\sin(3^\circ)+\sin(1^\circ)][\sin(3^\circ)-\sin(1^\circ)]} \\ &\quad+\frac{1}{[\sin(4^\circ)+\sin(1^\circ)][\sin(4^\circ)-\sin(1^\circ)]} \cdots \\ &\quad-\frac{1}{[\sin(89^\circ)+\sin(1^\circ)][\sin(89^\circ)-\sin(1^\circ)]} \end{align}$$

Now, I think I have to apply $\sin(90^\circ-\theta) = \cos(\theta)$, but I'm not very sure about where and how to apply this.

Tell me if I have to. And if so, where. All solutions are encouraged. :)

Also, the answer given was $\dfrac{\cot(2^\circ)}{\sin(2^\circ)}$.

Tell me if this is correct. If so, how do you reach to that answer?

Thank you.

Blue
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Tejas parashar
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1 Answers1

2

$$\begin{aligned} \frac1{\sin^2A-\sin^2B}&=\frac1{\sin (A+B)\sin(A-B)}\\ &=\underbrace{\frac{\sin(A+B)\cos(A-B)-\cos(A+B)\sin(A-B)}{\sin2B}}_{=1}\frac1{\sin (A+B)\sin(A-B)}\\ &=\frac{\cot(A-B)-\cot(A+B)}{\sin2B} \end{aligned}$$

so that with $A=k+1$, $B=1$ (in degrees) one obtains: $$\begin{aligned} &\sum_{k=1}^{88}\frac{(-1)^{k+1}}{\sin^2A-\sin^2B}=\frac1{\sin2}\sum_{k=1}^{88}(-1)^{k+1}\left[\cot k-\cot(k+2)\right]\\ &=\frac{1}{\sin2}[(\cot1-\cot3)-(\cot2-\cot4)+\cdots+(\cot87-\cot89)-(\cot88-\cot90)]\\ &=\frac{1}{\sin2}[\cot1-\cot2-\cot89+\cot90]=\frac{\cot2}{\sin2}, \end{aligned}$$ where in the last equality we used the identity $\cot x-\tan x=2\cot2x$.

user
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