find the range of function `f(x)=sin(x+(pi)/(6))+cos(x-(pi)/(6))`

Options

A. `[-sqrt(2), sqrt(2)]`
B. `[-sqrt(2)(sqrt(3)+1), sqrt(2)(sqrt(3)+1)]`
C. `[-(sqrt(3)+1)/(sqrt(2)), (sqrt(3)+1)/(sqrt(2))]`
D. `[-(sqrt(3)-1)/(sqrt(2)), (sqrt(3)-1)/(sqrt(2))]`

Answer

To find the range of the function \( f(x) = \sin\left(x + \frac{\pi}{6}\right) + \cos\left(x - \frac{\pi}{6}\right) \), we can use trigonometric identities and properties of sine and cosine functions. Here is a step-by-step solution:

### Step 1: Rewrite the function using trigonometric identities
Using the angle addition and subtraction formulas, we can rewrite the function:
\[ f(x) = \sin\left(x + \frac{\pi}{6}\right) + \cos\left(x - \frac{\pi}{6}\right) \]
\[ = \sin x \cos\frac{\pi}{6} + \cos x \sin\frac{\pi}{6} + \cos x \cos\frac{\pi}{6} + \sin x \sin\frac{\pi}{6} \]

### Step 2: Substitute the values of \(\cos\frac{\pi}{6}\) and \(\sin\frac{\pi}{6}\)
We know that:
\[ \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin\frac{\pi}{6} = \frac{1}{2} \]
Substituting these values into the equation gives:
\[ f(x) = \sin x \cdot \frac{\sqrt{3}}{2} + \cos x \cdot \frac{1}{2} + \cos x \cdot \frac{\sqrt{3}}{2} + \sin x \cdot \frac{1}{2} \]
\[ = \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) \sin x + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) \cos x \]

### Step 3: Combine the terms
Combining the coefficients:
\[ f(x) = \left(\frac{\sqrt{3} + 1}{2}\right) \sin x + \left(\frac{\sqrt{3} + 1}{2}\right) \cos x \]
This can be factored out:
\[ f(x) = \frac{\sqrt{3} + 1}{2} (\sin x + \cos x) \]

### Step 4: Find the range of \(\sin x + \cos x\)
The expression \(\sin x + \cos x\) can be rewritten using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The range of \(\sin\) function is \([-1, 1]\), thus:
\[ \sin x + \cos x \in [-\sqrt{2}, \sqrt{2}] \]

### Step 5: Scale the range
Now, we scale the range by \(\frac{\sqrt{3} + 1}{2}\):
\[ f(x) \in \left[-\frac{\sqrt{3} + 1}{2} \cdot \sqrt{2}, \frac{\sqrt{3} + 1}{2} \cdot \sqrt{2}\right] \]

### Step 6: Simplify the range
Calculating the endpoints:
\[ -\frac{\sqrt{3} + 1}{2} \cdot \sqrt{2} \quad \text{and} \quad \frac{\sqrt{3} + 1}{2} \cdot \sqrt{2} \]

Thus, the range of the function \( f(x) \) is:
\[ \left[-\frac{(\sqrt{3} + 1)\sqrt{2}}{2}, \frac{(\sqrt{3} + 1)\sqrt{2}}{2}\right] \]

### Final Answer
The range of the function \( f(x) = \sin\left(x + \frac{\pi}{6}\right) + \cos\left(x - \frac{\pi}{6}\right) \) is:
\[ \left[-\frac{(\sqrt{3} + 1)\sqrt{2}}{2}, \frac{(\sqrt{3} + 1)\sqrt{2}}{2}\right] \]