Find the rang of `f(x)=log_sqrt5(3+cos((3pi)/4+x)+cos((pi)/4+x)+cos((pi)/4-x)-cos((3pi)/4-x))`

Options

A. `(0,sqrt5)`
B. `[1/sqrt5,sqrt5]`
C. `[-2,2]`
D. `[0,2]`

Answer

To find the range of the function

\[ f(x) = \log_{\sqrt{5}} \left( 3 + \cos\left(\frac{3\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) - \cos\left(\frac{3\pi}{4} - x\right) \right), \]

we will simplify the expression inside the logarithm step by step.

### Step 1: Simplify the cosine terms

We start by rewriting the function:

\[ f(x) = \log_{\sqrt{5}} \left( 3 + \cos\left(\frac{3\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) - \cos\left(\frac{3\pi}{4} - x\right) \right). \]

Using the cosine addition and subtraction formulas:

1. \(\cos(a + b) = \cos a \cos b - \sin a \sin b\)
2. \(\cos(a - b) = \cos a \cos b + \sin a \sin b\)

We can group the terms appropriately.

### Step 2: Grouping and using identities

Now we can group the terms:

\[ \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) + \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right). \]

Using the cosine subtraction formula:

\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right), \]

we can simplify \(\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right)\) to:

\[ -2 \sin\left(\frac{3\pi/4 + 3\pi/4}{2}\right) \sin\left(\frac{2x}{2}\right) = -2 \sin\left(\frac{3\pi}{4}\right) \sin(x) = -2 \cdot \frac{1}{\sqrt{2}} \sin(x) = -\sqrt{2} \sin(x). \]

For the other two cosine terms, we can use the cosine addition formula:

\[ \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) = 2 \cos\left(\frac{\pi/4 + \pi/4}{2}\right) \cos\left(\frac{2x}{2}\right) = 2 \cos\left(\frac{\pi}{4}\right) \cos(x) = \sqrt{2} \cos(x). \]

### Step 3: Combine the results

Now we can substitute back into our function:

\[ f(x) = \log_{\sqrt{5}} \left( 3 - \sqrt{2} \sin(x) + \sqrt{2} \cos(x) \right). \]

### Step 4: Express in a single trigonometric function

We can rewrite \(-\sqrt{2} \sin(x) + \sqrt{2} \cos(x)\) as:

\[ \sqrt{2} \left( \cos(x) - \sin(x) \right). \]

The expression \(\cos(x) - \sin(x)\) can be rewritten as:

\[ \sqrt{2} \cos\left(x + \frac{\pi}{4}\right). \]

Thus, we have:

\[ f(x) = \log_{\sqrt{5}} \left( 3 + 2 \cos\left(x + \frac{\pi}{4}\right) \right). \]

### Step 5: Determine the range of the cosine function

The term \(2 \cos\left(x + \frac{\pi}{4}\right)\) varies from \(-2\) to \(2\). Therefore, the expression \(3 + 2 \cos\left(x + \frac{\pi}{4}\right)\) varies from \(3 - 2 = 1\) to \(3 + 2 = 5\).

### Step 6: Find the range of the logarithm

Now we can find the range of \(f(x)\):

\[ f(x) = \log_{\sqrt{5}}(y) \quad \text{where } y \in [1, 5]. \]

Calculating the logarithm:

- When \(y = 1\), \(f(x) = \log_{\sqrt{5}}(1) = 0\).
- When \(y = 5\), \(f(x) = \log_{\sqrt{5}}(5) = 2\).

### Conclusion

Thus, the range of \(f(x)\) is:

\[ \boxed{[0, 2]}. \]