The maximum value of `1+sin((pi)/(6)+theta)+2cos((pi)/(3)-theta)` for real values of `theta` is

Options

A. 3
B. 5
C. 4
D. none of these

Explanation

To find the maximum value of the expression \(1 + \sin\left(\frac{\pi}{6} + \theta\right) + 2\cos\left(\frac{\pi}{3} - \theta\right)\), we can follow these steps:

### Step 1: Rewrite the Expression
We start by rewriting the expression:
\[ f(\theta) = 1 + \sin\left(\frac{\pi}{6} + \theta\right) + 2\cos\left(\frac{\pi}{3} - \theta\right) \]

### Step 2: Use Trigonometric Identities
We know that:
- \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
- \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)

Using these values, we can express \(f(\theta)\) as:
\[ f(\theta) = 1 + \sin\left(\frac{\pi}{6} + \theta\right) + 2\left(\frac{1}{2}\cos(-\theta)\right) \]
This simplifies to:
\[ f(\theta) = 1 + \sin\left(\frac{\pi}{6} + \theta\right) + \cos(-\theta) \]
Since \(\cos(-\theta) = \cos(\theta)\), we have:
\[ f(\theta) = 1 + \sin\left(\frac{\pi}{6} + \theta\right) + \cos(\theta) \]

### Step 3: Find the Maximum Value of the Sine and Cosine Functions
To maximize \(f(\theta)\), we need to consider the maximum values of \(\sin\) and \(\cos\):
- The maximum value of \(\sin\) is 1.
- The maximum value of \(\cos\) is also 1.

### Step 4: Substitute Maximum Values
The maximum value of \(\sin\left(\frac{\pi}{6} + \theta\right)\) occurs when \(\frac{\pi}{6} + \theta = \frac{\pi}{2}\), leading to:
\[ \sin\left(\frac{\pi}{6} + \theta\right) = 1 \]
The maximum value of \(\cos(\theta)\) occurs when \(\theta = 0\), leading to:
\[ \cos(\theta) = 1 \]

### Step 5: Calculate the Maximum Value of \(f(\theta)\)
Substituting these maximum values into \(f(\theta)\):
\[ f(\theta)_{\text{max}} = 1 + 1 + 1 = 3 \]

### Step 6: Final Calculation
However, we need to consider that the maximum value of \(2\cos\left(\frac{\pi}{3} - \theta\right)\) can also contribute to the overall maximum. The maximum value of \(2\cos\left(\frac{\pi}{3} - \theta\right)\) is \(2\) when \(\theta = \frac{\pi}{3}\).

Thus, the maximum value of the entire expression is:
\[ f(\theta)_{\text{max}} = 1 + 1 + 2 = 4 \]

### Conclusion
The maximum value of \(1 + \sin\left(\frac{\pi}{6} + \theta\right) + 2\cos\left(\frac{\pi}{3} - \theta\right)\) is \(4\).

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