If ∑_{r=1}^{13} { 1 / (sin(π/4 + (r-1)π/6) sin(π/4 + rπ/6)) } = a√3 + b, a, b ∈ Z, then a² + b² is e

Q: Why do we multiply by $\sin(\frac{\pi}{6})$? Isn't that changing the value of the expression?

Great observation! We aren't changing the value; we are multiplying by $1$ in a clever disguise. Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, we are effectively multiplying by $\frac{1/2}{1/2} = 1$. This 'clever one' is the key to unlocking the telescoping structure.

Q: I got lost in the telescoping part. How do I know which terms cancel?

Think of it like a line of dominoes. When you write out the sum, you get $(\cot A_1 - \cot A_2) + (\cot A_2 - \cot A_3) + \dots + (\cot A_{13} - \cot A_{14})$. Notice that the second part of the first bracket cancels the first part of the second bracket. This chain reaction leaves only the very first and very last terms standing.

Q: How do I calculate $\cot(75^\circ)$ quickly?

You can derive it using $\cot(45^\circ + 30^\circ) = \frac{\cot 45^\circ \cot 30^\circ - 1}{\cot 45^\circ + \cot 30^\circ}$. Substituting the values, you get $\frac{1 \cdot \sqrt{3} - 1}{1 + \sqrt{3}}$. Rationalizing this gives $2 - \sqrt{3}$. Memorizing standard values like $\tan(75^\circ) = 2+\sqrt{3}$ and $\cot(75^\circ) = 2-\sqrt{3}$ is a huge time-saver for JEE.

Question

If $\sum_{r=1}^{13}\{\frac{1}{\sin(\frac{\pi}{4}+(r-1)\frac{\pi}{6})\sin(\frac{\pi}{4}+\frac{r\pi}{6})}\}=a\sqrt{3}+b,$ a, $b\in \mathbb{Z},$ then $a^{2}+b^{2}$ is equal to :

Options:

(A) 10

(B) 4

(C) 2

(D) 8

Sigma Percentile · Accepted Answer

### The Art of the Telescoping Series: A Journey Through Trigonometry

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of sines and angles. You see a summation from $r=1$ to $13$, and the denominator is a product of two sine functions.

Your instinct might be to panic, to try and expand everything, or to look for a complex identity. But stop. Take a breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask.

### Phase 1: Decoding the Anatomy

First, let us look at the angles. We define $A_r = \frac{\pi}{4} + (r-1)\frac{\pi}{6}$. The term following it is $A_{r+1} = \frac{\pi}{4} + r\frac{\pi}{6}$.

Now, calculate the difference:

$$A_{r+1} - A_r = \left( \frac{\pi}{4} + \frac{r\pi}{6} \right) - \left( \frac{\pi}{4} + \frac{(r-1)\pi}{6} \right)$$

The $\frac{\pi}{4}$ terms vanish, and we are left with:

$$A_{r+1} - A_r = \frac{r\pi}{6} - \frac{(r-1)\pi}{6} = \frac{\pi}{6}$$

This is the heartbeat of the problem. The difference is constant! Whenever you see a product of trigonometric functions in the denominator, and the difference between the angles is constant, you are dealing with a telescoping series.

### Phase 2: The Telescoping Magic

We need to transform our general term $T_r = \frac{1}{\sin(A_r)\sin(A_{r+1})}$ into something that can collapse. We use the "clever one" trick. We multiply and divide by $\sin(\frac{\pi}{6})$.

Because $\sin(\frac{\pi}{6}) = \sin(A_{r+1} - A_r)$, this allows us to rewrite the numerator using the sine subtraction identity: $\sin(A-B) = \sin A \cos B - \cos A \sin B$.

So, our term becomes:

$$T_r = \frac{1}{\sin(\frac{\pi}{6})} \cdot \frac{\sin(A_{r+1} - A_r)}{\sin A_r \sin A_{r+1}}$$

Expanding the numerator, we get:

$$T_r = 2 \left[ \frac{\sin A_{r+1} \cos A_r - \cos A_{r+1} \sin A_r}{\sin A_r \sin A_{r+1}} \right]$$

When we split this fraction, the sine terms cancel out beautifully, leaving us with:

$$T_r = 2 ( \cot A_r - \cot A_{r+1} )$$

### Phase 3: The Massacre

Now, we sum this from $r=1$ to $13$. The expression becomes:

$$S = 2 \sum_{r=1}^{13} ( \cot A_r - \cot A_{r+1} )$$

Let us expand this sum:

$$S = 2 [ (\cot A_1 - \cot A_2) + (\cot A_2 - \cot A_3) + (\cot A_3 - \cot A_4) + \dots + (\cot A_{13} - \cot A_{14}) ]$$

Look at the terms! The $-\cot A_2$ from the first bracket is annihilated by the $+\cot A_2$ from the second. This chain reaction continues, consuming every intermediate term.

This is the "telescoping" effect—the series collapses like a folding telescope, leaving only the first and the last terms standing:

$$S = 2 [ \cot A_1 - \cot A_{14} ]$$

### Phase 4: The Final Tally

We are almost there. We just need to evaluate the boundary angles. For $r=1$, $A_1 = \frac{\pi}{4}$.

For $r=14$:

$$A_{14} = \frac{\pi}{4} + 13\frac{\pi}{6} = \frac{\pi}{4} + 2\pi + \frac{\pi}{6} = 2\pi + \frac{5\pi}{12}$$

We know $\cot(\frac{\pi}{4}) = 1$. For $\cot(2\pi + \frac{5\pi}{12})$, the $2\pi$ is just a full rotation, so we are looking at $\cot(75^\circ)$, which is $2 - \sqrt{3}$.

Substituting these back:

$$S = 2 [ 1 - (2 - \sqrt{3}) ] = 2 [ \sqrt{3} - 1 ] = 2\sqrt{3} - 2$$

Comparing this to $a\sqrt{3} + b$, we find $a=2$ and $b=-2$. The final answer, **$a^2 + b^2$**, is:

$$2^2 + (-2)^2 = 4 + 4 = 8$$

You have conquered the beast. Remember, in JEE, it is not about brute force; it is about finding the pattern and letting the math do the work for you.

Animated Solution for Mathematics - Trigonometry: If $\sum_{r = 1}^{13} {\frac{1}{s i n ( \frac{π}{4} + ( r - 1 ) \frac{π}{6} ) s i n ( \frac{π}{4} + \frac{r π}{6} )}} = a 3 + b,$ a, $b \in Z,$ then $a^{2} + b^{2}$ is equal to :

Select Answer:

Visualized Solution

Identifying the General Term

The Constant Difference Trick

Manipulating the Numerator

Applying Sine Subtraction Identity

Converting to Cotangent

The Summation and Telescoping Effect

Cancelling Intermediate Terms

Final Surviving Terms

Evaluating the Angles

Calculating Cotangent Values

Final Sum Calculation

Finding a and b

The Sigma Insight: Trigonometric Ratios and Identities

The Art of the Telescoping Series

Phase 1

Phase 2

Phase 3

Phase 4

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