Key Concepts and Formulas

1. Trigonometric Identities: We will utilize the sum-to-product and angle addition/subtraction formulas. Specifically,
   • $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$
   • $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$
   • $\cos(\pi - \theta) = -\cos\theta$ and $\sin(\pi - \theta) = \sin\theta$.
2. Range of Basic Trigonometric Functions: The range of $\cos x$ and $\sin x$ is $[-1, 1]$ for all real $x$.
3. Properties of Logarithms: For a base $b > 1$, the logarithmic function $\log_b y$ is an increasing function. This means if $y_1 \le y_2$, then $\log_b y_1 \le \log_b y_2$. Also, $\log_b 1 = 0$.

Step-by-Step Solution

Step 1: Simplify the Argument of the Logarithm Let the argument of the logarithm be $g(x)$. $g(x) = 3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)$ We will group terms strategically to apply trigonometric identities. $g(x) = 3 + \left[ \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) \right] + \left[ \cos \left( {{{3\pi } \over 4} + x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right) \right]$

Step 2: Apply Sum-to-Product Formulas For the first bracket, using $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$: Let $A = {\pi \over 4} + x$ and $B = {\pi \over 4} - x$. $\frac{A+B}{2} = \frac{{\pi \over 4} + x + {\pi \over 4} - x}{2} = \frac{{\pi \over 2}}{2} = \frac{\pi}{4}$. $\frac{A-B}{2} = \frac{{\pi \over 4} + x - ({\pi \over 4} - x)}{2} = \frac{2x}{2} = x$. So, $\cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) = 2\cos \left( {{\pi \over 4}} \right)\cos (x)$.

For the second bracket, using $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$: Let $A = {{{3\pi } \over 4} + x}$ and $B = {{{3\pi } \over 4} - x}$. $\frac{A+B}{2} = \frac{{{{3\pi } \over 4} + x + {{3\pi } \over 4} - x}}{2} = \frac{{3\pi \over 2}}{2} = \frac{3\pi}{4}$. $\frac{A-B}{2} = \frac{{{{3\pi } \over 4} + x - ({{3\pi } \over 4} - x)}}{2} = \frac{2x}{2} = x$. So, $\cos \left( {{{3\pi } \over 4} + x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right) = -2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)$.

Substituting these back into $g(x)$: $g(x) = 3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)$

Step 3: Evaluate Known Trigonometric Values and Substitute We know that $\cos \left( {{\pi \over 4}} \right) = \frac{1}{\sqrt{2}}$. For $\sin \left( {{{3\pi } \over 4}} \right)$, we use the identity $\sin(\pi - \theta) = \sin\theta$: $\sin \left( {{{3\pi } \over 4}} \right) = \sin \left( {\pi - {\pi \over 4}} \right) = \sin \left( {{\pi \over 4}} \right) = \frac{1}{\sqrt{2}}$.

Substitute these values into the expression for $g(x)$: $g(x) = 3 + 2 \left( \frac{1}{\sqrt{2}} \right) \cos (x) - 2 \left( \frac{1}{\sqrt{2}} \right) \sin (x)$ $g(x) = 3 + \sqrt{2} \cos (x) - \sqrt{2} \sin (x)$ $g(x) = 3 + \sqrt{2} (\cos (x) - \sin (x))$

Step 4: Determine the Range of the Simplified Trigonometric Expression We need to find the range of $\cos (x) - \sin (x)$. This expression is of the form $a\cos x + b\sin x$, where $a=1$ and $b=-1$. The range of $a\cos x + b\sin x$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$. Here, $\sqrt{a^2+b^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$. So, the range of $\cos (x) - \sin (x)$ is $[-\sqrt{2}, \sqrt{2}]$.

Now, we find the range of $\sqrt{2} (\cos (x) - \sin (x))$. Multiplying the range by $\sqrt{2}$: $\sqrt{2} \times [-\sqrt{2}, \sqrt{2}] = [-\sqrt{2} \times \sqrt{2}, \sqrt{2} \times \sqrt{2}] = [-2, 2]$.

Finally, we find the range of $g(x) = 3 + \sqrt{2} (\cos (x) - \sin (x))$. We add 3 to the range $[-2, 2]$: Range of $g(x)$ is $[3 + (-2), 3 + 2] = [1, 5]$.

Step 5: Determine the Range of the Logarithmic Function The function is $f(x) = {\log _{\sqrt 5 }}\left( {g(x)} \right)$. The base of the logarithm is $\sqrt{5}$, which is greater than 1. Therefore, the logarithmic function is strictly increasing. We found that the range of $g(x)$ is $[1, 5]$. Since the logarithm is increasing, we apply it to the range of $g(x)$: Range of $f(x) = \left[ {\log _{\sqrt 5 }} (1), {\log _{\sqrt 5 }} (5) \right]$.

We evaluate the endpoints: $\log _{\sqrt 5 } (1) = 0$ (since any base raised to the power of 0 is 1). To evaluate $\log _{\sqrt 5 } (5)$, we can set it equal to $y$: $\log _{\sqrt 5 } (5) = y \implies (\sqrt{5})^y = 5 \implies (5^{1/2})^y = 5^1 \implies 5^{y/2} = 5^1 \implies \frac{y}{2} = 1 \implies y = 2$. So, $\log _{\sqrt 5 } (5) = 2$.

Therefore, the range of $f(x)$ is $[0, 2]$.

Common Mistakes & Tips

• Incorrect application of sum-to-product formulas: Ensure the correct formula is used for sum and difference of cosines.
• Sign errors: Pay close attention to the signs when applying formulas, especially the $\cos A - \cos B$ formula.
• Logarithm base: Remember that the behavior of a logarithmic function (increasing or decreasing) depends on its base. $\sqrt{5} > 1$, so it's increasing.
• Range of trigonometric expressions: Always recall that $a\cos\theta + b\sin\theta$ has a range of $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.

Summary

To find the range of the given logarithmic function, we first simplified the argument of the logarithm by applying trigonometric sum-to-product identities and evaluating known trigonometric values. This resulted in a simplified expression $g(x) = 3 + \sqrt{2}(\cos x - \sin x)$. We then determined the range of this expression by finding the range of $\cos x - \sin x$ and then manipulating it. The range of $g(x)$ was found to be $[1, 5]$. Finally, since the base of the logarithm, $\sqrt{5}$, is greater than 1, the logarithmic function is increasing. Applying the logarithm to the range of $g(x)$ yielded the range of $f(x)$ as $[0, 2]$.

The final answer is $\boxed{[0, 2]}$.