The maximum values of `3 costheta+5sin(theta-(pi)/(6))` for any real value of `theta` is:

Options

A. `sqrt(1 9)`
B. `sqrt(79)/(2)`
C. `sqrt(31)`
D. `sqrt(34)`

Answer

To find the maximum value of the expression \(3 \cos \theta + 5 \sin \left(\theta - \frac{\pi}{6}\right)\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = 3 \cos \theta + 5 \sin \left(\theta - \frac{\pi}{6}\right) \]
Using the sine subtraction formula, we can rewrite \( \sin \left(\theta - \frac{\pi}{6}\right) \) as:
\[ \sin \left(\theta - \frac{\pi}{6}\right) = \sin \theta \cos \frac{\pi}{6} - \cos \theta \sin \frac{\pi}{6} \]
Substituting the values of \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \), we have:
\[ \sin \left(\theta - \frac{\pi}{6}\right) = \sin \theta \cdot \frac{\sqrt{3}}{2} - \cos \theta \cdot \frac{1}{2} \]

### Step 2: Substitute back into the expression
Now substituting this back into our expression for \(y\):
\[ y = 3 \cos \theta + 5 \left(\frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta\right) \]
This simplifies to:
\[ y = 3 \cos \theta + \frac{5\sqrt{3}}{2} \sin \theta - \frac{5}{2} \cos \theta \]
Combining the cosine terms:
\[ y = \left(3 - \frac{5}{2}\right) \cos \theta + \frac{5\sqrt{3}}{2} \sin \theta \]
\[ y = \frac{1}{2} \cos \theta + \frac{5\sqrt{3}}{2} \sin \theta \]

### Step 3: Find the maximum value
The maximum value of \(a \cos \theta + b \sin \theta\) is given by \(\sqrt{a^2 + b^2}\). Here, \(a = \frac{1}{2}\) and \(b = \frac{5\sqrt{3}}{2}\).

Calculating \(a^2 + b^2\):
\[ a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ b^2 = \left(\frac{5\sqrt{3}}{2}\right)^2 = \frac{75}{4} \]
Adding these together:
\[ a^2 + b^2 = \frac{1}{4} + \frac{75}{4} = \frac{76}{4} = 19 \]

Thus, the maximum value of \(y\) is:
\[ \sqrt{19} \]

### Final Answer
The maximum value of \(3 \cos \theta + 5 \sin \left(\theta - \frac{\pi}{6}\right)\) is:
\[ \sqrt{76} = 2\sqrt{19} \]