Question

Let $f: \mathbb{R} \to \mathbb{R}$ 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 :

• A. $\sec^2(1) \sec(21) \cos(20)$
• B. $\text{cosec}^2(21) \cos(20) \cos(2)$
• C. $\text{cosec}(1) \text{cosec}(21) \sin(20)$
• D. $\sec(21) \sin(20) \sin(2)$

Hint:

When dealing with trigonometric summations, always look for an opportunity to create a telescoping series, often using sum-to-product or difference identities. If a direct approach doesn't match the options, re-examine the question for potential typos that might simplify the expression.

Answer 1

The Correct Option is C

The functional equation \[ f(x+y)+f(x-y)=2f(x)f(y) \] has the standard solution \[ f(x)=\cos(ax) \] Using $f(\tfrac12)=-1$: \[ \cos\!\left(\tfrac a2\right)=-1 \Rightarrow \tfrac a2=\pi \Rightarrow a=2\pi \] Hence, \[ f(x)=\cos(2\pi x) \] For integer $k$, \[ f(k)=\cos(2\pi k)=1 \] So the sum becomes: \[ \sum_{k=1}^{20}\frac{1}{\sin k\,\sin(k+1)} \] Using the identity: \[ \cot k-\cot(k+1)=\frac{\sin(1)}{\sin k\,\sin(k+1)} \] \[ \Rightarrow \frac{1}{\sin k\,\sin(k+1)} =\frac{1}{\sin(1)}\big[\cot k-\cot(k+1)\big] \] Hence the sum telescopes: \[ \sum_{k=1}^{20}\frac{1}{\sin k\,\sin(k+1)} =\frac{1}{\sin(1)}\big[\cot(1)-\cot(21)\big] \] \[ =\frac{1}{\sin(1)}\cdot \frac{\sin(21-1)}{\sin(1)\sin(21)} =\csc(1)\csc(21)\sin(20) \]