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Step-by-Step Solution
Step 1: Identify the Functional Form of f(x)
The given functional equation is
$ f(x + y) + f(x - y) = 2 f(x) f(y). $
A well-known class of solutions to such equations is
$ f(x) = \cos(\lambda x) \text{ or } f(x) = \cosh(\lambda x). $
Since we are working over real functions and the condition
$ f\!\bigl(\tfrac{1}{2}\bigr) = -1 $
suggests a cosine form (as cosh cannot take negative values for real arguments), we consider:
$ f(x) = \cos(\lambda x). $
Step 2: Use the Initial Condition f(1/2) = -1
Substitute
$ x = \tfrac{1}{2} $
into
$ f(x) = \cos(\lambda x): $
$ f\!\bigl(\tfrac{1}{2}\bigr) = \cos\!\Bigl(\lambda \cdot \tfrac{1}{2}\Bigr) = -1. $
Hence,
$ \cos\!\bigl(\tfrac{\lambda}{2}\bigr) = -1
\quad \Longrightarrow \quad
\tfrac{\lambda}{2} = \pi + 2\pi m,\ \text{for integer }m. $
Taking the principal solution for simplicity,
$ \tfrac{\lambda}{2} = \pi
\quad \Rightarrow \quad
\lambda = 2\pi. $
Therefore,
$ f(x) = \cos\!\bigl(2 \pi x\bigr). $
Step 3: Evaluate f(k) for Integer k
For a natural number
$ k, $
$ f(k) = \cos\!\Bigl(2\pi \cdot k\Bigr). $
Since
$ 2\pi k $
represents an integral multiple of
$ 2\pi, $
we have
$ \cos(2\pi k) = 1. $
Hence,
$ f(k) = 1 \quad \text{for all integer } k. $
Step 4: Rewrite the Required Sum
The question asks us to find the value of
$ \displaystyle \sum_{k=1}^{20} \frac{1}{\sin(k)\,\sin\!\bigl(k + f(k)\bigr)}. $
Since
$ f(k) = 1, $
we rewrite the sum as
$ \displaystyle \sum_{k=1}^{20} \frac{1}{\sin(k)\,\sin(k+1)}. $
Step 5: Use a Telescoping Trigonometric Identity
Recall the identity for the difference of cotangents:
$ \cot(A) - \cot(B)
= \frac{\sin\bigl(B - A\bigr)}{\sin(A)\,\sin(B)}. $
For
$ B = A + 1, $
it gives
$ \cot(A) - \cot(A+1)
= \frac{\sin\!\bigl((A+1)-A\bigr)}{\sin(A)\,\sin(A+1)}
= \frac{\sin(1)}{\sin(A)\,\sin(A+1)}.
$
Hence,
$ \frac{1}{\sin(A)\,\sin(A+1)}
= \frac{\cot(A) - \cot(A+1)}{\sin(1)}.
$
Applying this to each term in the sum, we get
$ \displaystyle
\sum_{k=1}^{20} \frac{1}{\sin(k)\,\sin(k+1)}
= \frac{1}{\sin(1)} \sum_{k=1}^{20} \Bigl[\cot(k) - \cot(k+1)\Bigr].
$
Step 6: Perform the Telescoping Summation
Observe that
$ \cot(k) - \cot(k+1) $
terms telescope when summing from
$ k=1 $
to
$ k=20. $
Hence, the sum simplifies as follows:
$ \displaystyle
\sum_{k=1}^{20} \Bigl[\cot(k) - \cot(k+1)\Bigr]
= \cot(1) - \cot(21).
$
So,
$ \displaystyle
\sum_{k=1}^{20} \frac{1}{\sin(k)\,\sin(k+1)}
= \frac{\cot(1) - \cot(21)}{\sin(1)}.
$
Step 7: Express the Result Appropriately
We want this final expression in the form of
$ \csc^2(1)\,\csc(21)\,\sin(20). $
Using trigonometric manipulations and sum-to-product (or known transformations), one can show that
$ \frac{\cot(1) - \cot(21)}{\sin(1)}
= \csc^2(1)\,\csc(21)\,\sin(20).
$
Thus, the value of the given sum is
$ \csc^2(1)\,\csc(21)\,\sin(20). $
Final Answer
$ \displaystyle \csc^2(1)\,\csc(21)\,\sin(20).
