Find `max^(m)&min^(m)` value of `cos^(2)theta+sintheta`

Answer

To find the maximum and minimum values of the function \( f(\theta) = \cos^2 \theta + \sin \theta \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting \( \cos^2 \theta \) in terms of \( \sin \theta \):
\[ f(\theta) = \cos^2 \theta + \sin \theta = 1 - \sin^2 \theta + \sin \theta \]
This simplifies to:
\[ f(\theta) = 1 - \sin^2 \theta + \sin \theta \]
Thus, we can express it as:
\[ f(\theta) = -\sin^2 \theta + \sin \theta + 1 \]

### Step 2: Completing the square
Next, we will complete the square for the quadratic expression in \( \sin \theta \):
\[ f(\theta) = -(\sin^2 \theta - \sin \theta) + 1 \]
To complete the square, we take the coefficient of \( \sin \theta \) (which is -1), halve it, and square it:
\[ \left(\frac{-1}{2}\right)^2 = \frac{1}{4} \]
Now, we can rewrite the expression:
\[ f(\theta) = -\left(\sin^2 \theta - \sin \theta + \frac{1}{4} - \frac{1}{4}\right) + 1 \]
This simplifies to:
\[ f(\theta) = -\left(\left(\sin \theta - \frac{1}{2}\right)^2 - \frac{1}{4}\right) + 1 \]
\[ f(\theta) = -\left(\sin \theta - \frac{1}{2}\right)^2 + \frac{1}{4} + 1 \]
\[ f(\theta) = -\left(\sin \theta - \frac{1}{2}\right)^2 + \frac{5}{4} \]

### Step 3: Finding maximum and minimum values
Now we analyze the expression:
\[ f(\theta) = -\left(\sin \theta - \frac{1}{2}\right)^2 + \frac{5}{4} \]
The term \(-\left(\sin \theta - \frac{1}{2}\right)^2\) reaches its maximum value of 0 when \(\sin \theta = \frac{1}{2}\). Thus, the maximum value of \( f(\theta) \) is:
\[ f_{\text{max}} = 0 + \frac{5}{4} = \frac{5}{4} \]

For the minimum value, \(-\left(\sin \theta - \frac{1}{2}\right)^2\) reaches its minimum when \(\sin \theta\) is at its extreme values, either -1 or 1. Let's evaluate both cases:
1. When \(\sin \theta = -1\):
\[ f(-1) = -\left(-1 - \frac{1}{2}\right)^2 + \frac{5}{4} = -\left(-\frac{3}{2}\right)^2 + \frac{5}{4} = -\frac{9}{4} + \frac{5}{4} = -1 \]
2. When \(\sin \theta = 1\):
\[ f(1) = -\left(1 - \frac{1}{2}\right)^2 + \frac{5}{4} = -\left(\frac{1}{2}\right)^2 + \frac{5}{4} = -\frac{1}{4} + \frac{5}{4} = 1 \]

Thus, the minimum value of \( f(\theta) \) is:
\[ f_{\text{min}} = -1 \]

### Final Result
The maximum value of \( f(\theta) = \cos^2 \theta + \sin \theta \) is \( \frac{5}{4} \) and the minimum value is \( -1 \).