A continuous, even periodic function f with period 8 is such that `f(0)=0,f(1)=-2,f(2)=1,f(3)=2,f(4)=3,` then the value of `tan^(-1) tan{f(-5)+f(20)+cos^(-1)(f(-10))+f(17)}` is equal to

Options

A. `2pi-3`
B. `3-2pi`
C. `2pi+3`
D. `3-pi`

Explanation

To solve the problem step by step, we will use the properties of the periodic function \( f \) and the values given in the question.

### Step 1: Identify the periodic function properties
Given that \( f \) is a continuous, even periodic function with a period of 8, we know that:
\[ f(x) = f(x + 8) \]
for any \( x \).

### Step 2: Calculate \( f(-5) \)
Using the periodic property:
\[ f(-5) = f(-5 + 8) = f(3) \]
From the given values, \( f(3) = 2 \). Thus:
\[ f(-5) = 2 \]

### Step 3: Calculate \( f(20) \)
Using the periodic property:
\[ f(20) = f(20 - 16) = f(4) \]
From the given values, \( f(4) = 3 \). Thus:
\[ f(20) = 3 \]

### Step 4: Calculate \( f(-10) \)
Using the periodic property:
\[ f(-10) = f(-10 + 16) = f(6) \]
Since \( f \) is an even function:
\[ f(6) = f(-2) \]
Using the periodic property again:
\[ f(-2) = f(-2 + 8) = f(6) \]
Now, we need to find \( f(2) \) since \( f(-2) = f(2) \) and from the given values, \( f(2) = 1 \). Thus:
\[ f(-10) = 1 \]

### Step 5: Calculate \( f(17) \)
Using the periodic property:
\[ f(17) = f(17 - 16) = f(1) \]
From the given values, \( f(1) = -2 \). Thus:
\[ f(17) = -2 \]

### Step 6: Substitute values into the expression
Now we substitute the calculated values into the expression:
\[ f(-5) + f(20) + \cos^{-1}(f(-10)) + f(17) \]
Substituting the values we found:
\[ = 2 + 3 + \cos^{-1}(1) - 2 \]
Since \( \cos^{-1}(1) = 0 \):
\[ = 2 + 3 + 0 - 2 = 3 \]

### Step 7: Calculate \( \tan^{-1}(\tan(3)) \)
The expression simplifies to:
\[ \tan^{-1}(\tan(3)) \]
Since \( 3 \) is within the range of \( \tan^{-1} \), we have:
\[ \tan^{-1}(\tan(3)) = 3 \]

### Final Answer
Thus, the value of \( \tan^{-1} \tan(f(-5) + f(20) + \cos^{-1}(f(-10)) + f(17)) \) is:
\[ \boxed{3} \]