Let f : R → R be defined as f(x + y) + f(x - y) = 2f(x)f(y), f(1/2) = -1. Then the value of Σ(k=1 to

Q: Wait, why do we assume $f(x) = \cos(\lambda x)$? Couldn't it be something else?

Great question! D'Alembert's functional equation $f(x+y) + f(x-y) = 2f(x)f(y)$ is a foundational result. For continuous functions, the only solutions are $f(x) = \cos(\lambda x)$ or $f(x) = \cosh(\lambda x)$. Given the context of the problem involving sine functions in the summation, the cosine form is the natural fit.

Q: Why does $f(k)$ become $1$ for all integers $k$?

Excellent observation. We found $f(x) = \cos(2\pi x)$. When $k$ is an integer, $2\pi k$ is an even multiple of $\pi$. Since $\cos(2n\pi) = 1$ for any integer $n$, $f(k)$ collapses to $1$ for all $k \in \mathbb{N}$. It simplifies the problem beautifully!

Q: Why multiply by $\sin(1)$? How would I ever think of that?

That is the 'telescoping trick.' When you see a product of sines in the denominator, $\sin(k)\sin(k+1)$, you want to create a difference in the numerator. Since $(k+1) - k = 1$, multiplying by $\sin(1)$ allows you to use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ to split the fraction into two parts that cancel out. It's a standard JEE technique for series summation!

Question

Let $f : \mathbf{R} 	o \mathbf{R}$ be defined as $f(x + y) + f(x - y) = 2f(x)f(y)$, $f(\frac{1}{2}) = -1$. Then the value of $\sum_{k=1}^{20} \frac{1}{\sin(k)\sin(k+f(k))}$ is equal to:

Options:

(A) $	ext{cosec}^2(21) \cos(20) \cos(2)$

(B) $\sec^2(1) \sec(21) \cos(20)$

(C) $	ext{cosec}^2(1) 	ext{cosec}(21) \sin(20)$

(D) $\sec^2(21) \sin(20) \sin(2)$

Sigma Percentile · Accepted Answer

### Decoding the Functional Equation

We start with the equation $f(x + y) + f(x - y) = 2f(x)f(y)$. This is the legendary **D'Alembert's functional equation**, which serves as the foundation for many trigonometric relationships.

When you encounter this structure, your mind should immediately jump to the cosine function. This is because the cosine addition formula,

$$ \cos(x+y) + \cos(x-y) = 2\cos(x)\cos(y) $$

perfectly mirrors the given equation. Thus, we assume the solution takes the form $f(x) = \cos(\lambda x)$.

### The Boundary Condition

To find the specific frequency $\lambda$, we utilize the given condition $f(\frac{1}{2}) = -1$. Substituting this into our assumed solution, we obtain:

$$ \cos\left(\lambda \cdot \frac{1}{2}ight) = -1 $$

We know that $\cos(	heta) = -1$ when $	heta = \pi$. Therefore, we set $\frac{\lambda}{2} = \pi$, which leads us to $\lambda = 2\pi$.

Our function is now locked in as:

$$ f(x) = \cos(2\pi x) $$

### The Integer Simplification

We now evaluate $f(k)$ for $k \in \{1, 2, \dots, 20\}$. Since $k$ is an integer, $f(k) = \cos(2\pi k)$.

Because $\cos(2n\pi) = 1$ for any integer $n$, $f(k)$ is always $1$. Consequently, the summation, which initially appeared intimidating, simplifies to:

$$ S = \sum_{k=1}^{20} \frac{1}{\sin(k)\sin(k+1)} $$

### The Telescoping Magic

To resolve the product of sines in the denominator, we employ the **telescoping trick**. We multiply and divide the general term by $\sin(1)$, noting that $1 = (k+1) - k$.

Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we rewrite the numerator as $\sin((k+1)-k) = \sin(k+1)\cos(k) - \cos(k+1)\sin(k)$. Dividing this by $\sin(k)\sin(k+1)$ splits the term into:

$$ \frac{1}{\sin(1)} [\cot(k) - \cot(k+1)] $$

### The Final Calculation

When we expand the sum, we get:

$$ S = \frac{1}{\sin(1)} [(\cot 1 - \cot 2) + (\cot 2 - \cot 3) + \dots + (\cot 20 - \cot 21)] $$

Everything in the middle vanishes, leaving us with:

$$ S = \frac{1}{\sin(1)} [\cot(1) - \cot(21)] $$

Converting back to sines and cosines, we have:

$$ S = \frac{1}{\sin(1)} \left[ \frac{\cos(1)}{\sin(1)} - \frac{\cos(21)}{\sin(21)} ight] = \frac{\sin(21-1)}{\sin^2(1)\sin(21)} = \frac{\sin(20)}{\sin^2(1)\sin(21)} $$

This simplifies to the final result:

$$ \mathbf{	ext{cosec}^2(1) 	ext{cosec}(21) \sin(20)} $$

Animated Solution for Mathematics - Trigonometry: Let $f : R \to R$ be defined as $f (x + y) + f (x - y) = 2 f (x) f (y)$ , $f (\frac{1}{2}) = - 1$ . Then the value of $\sum_{k = 1}^{20} \frac{1}{s i n ( k ) s i n ( k + f ( k ))}$ is equal to:

Select Answer:

Visualized Solution

Analyzing the Functional Equation

Applying the Boundary Condition

Determining the Function $f (x)$

Evaluating $f (k)$ for Natural Numbers

Setting up the Summation

The Telescoping Trick

Applying Sine Addition Formula

Simplifying to Cotangent Terms

Expanding the Summation

Result of Telescoping

Final Trigonometric Simplification

The Sigma Insight: Trigonometric Ratios and Identities

Decoding the Functional Equation

The Boundary Condition

The Integer Simplification

The Telescoping Magic

The Final Calculation

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