Minimum value of `3 sin theta+4 cos theta` in the interval `[0, (pi)/(2)]` is :

Options

A. -5
B. 3
C. 4
D. `(7)/(sqrt(2))`

Answer

To find the minimum value of the expression \(3 \sin \theta + 4 \cos \theta\) in the interval \([0, \frac{\pi}{2}]\), we can follow these steps:

### Step 1: Identify the expression
We want to minimize the expression:
\[ f(\theta) = 3 \sin \theta + 4 \cos \theta \]

### Step 2: Find the derivative
To find the critical points, we need to take the derivative of \(f(\theta)\) with respect to \(\theta\):
\[ f'(\theta) = 3 \cos \theta - 4 \sin \theta \]

### Step 3: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 3 \cos \theta - 4 \sin \theta = 0 \]
This can be rearranged to:
\[ 3 \cos \theta = 4 \sin \theta \]
Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ 3 = 4 \tan \theta \]
Thus,
\[ \tan \theta = \frac{3}{4} \]

### Step 4: Find \(\theta\)
Now, we can find \(\theta\) using the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 5: Check if \(\theta\) is in the interval
We need to check if \(\theta = \tan^{-1}\left(\frac{3}{4}\right)\) lies within the interval \([0, \frac{\pi}{2}]\). Since \(\tan^{-1}\left(\frac{3}{4}\right)\) is a positive angle, it is indeed in the interval.

### Step 6: Evaluate the function at the critical point and endpoints
Now we evaluate the function at the critical point and at the endpoints of the interval:
1. At \(\theta = 0\):
\[ f(0) = 3 \sin(0) + 4 \cos(0) = 0 + 4 = 4 \]
2. At \(\theta = \frac{\pi}{2}\):
\[ f\left(\frac{\pi}{2}\right) = 3 \sin\left(\frac{\pi}{2}\right) + 4 \cos\left(\frac{\pi}{2}\right) = 3 + 0 = 3 \]
3. At \(\theta = \tan^{-1}\left(\frac{3}{4}\right)\):
We can substitute \(\theta\) back into the original function:
\[ f\left(\tan^{-1}\left(\frac{3}{4}\right)\right) = 3 \sin\left(\tan^{-1}\left(\frac{3}{4}\right)\right) + 4 \cos\left(\tan^{-1}\left(\frac{3}{4}\right)\right) \]
Using the identity \(\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1+x^2}}\) and \(\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}}\):
\[ \sin\left(\tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{3}{5}, \quad \cos\left(\tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{4}{5} \]
Thus,
\[ f\left(\tan^{-1}\left(\frac{3}{4}\right)\right) = 3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = \frac{9}{5} + \frac{16}{5} = \frac{25}{5} = 5 \]

### Step 7: Determine the minimum value
Now we compare the values:
- \(f(0) = 4\)
- \(f\left(\frac{\pi}{2}\right) = 3\)
- \(f\left(\tan^{-1}\left(\frac{3}{4}\right)\right) = 5\)

The minimum value occurs at \(\theta = \frac{\pi}{2}\):
\[ \text{Minimum value} = 3 \]

### Final Answer
The minimum value of \(3 \sin \theta + 4 \cos \theta\) in the interval \([0, \frac{\pi}{2}]\) is **3**.