discuss the extremum of `f(theta)=sin^pthetacos^qtheta, p , q gt0,0ltthetaltpi/2`

Explanation

To discuss the extremum of the function \( f(\theta) = \sin^p \theta \cos^q \theta \) for \( p, q > 0 \) and \( 0 < \theta < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Differentiate the Function
We start by differentiating the function \( f(\theta) \) with respect to \( \theta \).

\[ f(\theta) = \sin^p \theta \cos^q \theta \]

Using the product rule and chain rule, we differentiate:

\[ f'(\theta) = \frac{d}{d\theta}(\sin^p \theta) \cdot \cos^q \theta + \sin^p \theta \cdot \frac{d}{d\theta}(\cos^q \theta) \]

Calculating each derivative separately:

1. \( \frac{d}{d\theta}(\sin^p \theta) = p \sin^{p-1} \theta \cos \theta \)
2. \( \frac{d}{d\theta}(\cos^q \theta) = -q \cos^{q-1} \theta \sin \theta \)

Combining these, we get:

\[ f'(\theta) = p \sin^{p-1} \theta \cos^q \theta - q \sin^p \theta \cos^{q-1} \theta \]

### Step 2: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:

\[ p \sin^{p-1} \theta \cos^q \theta - q \sin^p \theta \cos^{q-1} \theta = 0 \]

Rearranging gives:

\[ p \sin^{p-1} \theta \cos^q \theta = q \sin^p \theta \cos^{q-1} \theta \]

### Step 3: Factor the Equation
We can factor out common terms:

\[ \sin^{p-1} \theta \cos^{q-1} \theta \left( p \cos \theta - q \sin \theta \right) = 0 \]

This gives us two cases:

1. \( \sin^{p-1} \theta = 0 \) (which occurs at \( \theta = 0 \))
2. \( \cos^{q-1} \theta = 0 \) (which occurs at \( \theta = \frac{\pi}{2} \))
3. \( p \cos \theta - q \sin \theta = 0 \)

### Step 4: Solve for Critical Points
From the third equation, we can solve for \( \theta \):

\[ p \cos \theta = q \sin \theta \implies \tan \theta = \frac{p}{q} \]

Thus,

\[ \theta = \tan^{-1}\left(\frac{p}{q}\right) \]

### Step 5: Evaluate the Function at Critical Points
Now we evaluate \( f(\theta) \) at the critical points:

1. At \( \theta = 0 \):
\[ f(0) = \sin^p(0) \cos^q(0) = 0 \]

2. At \( \theta = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin^p\left(\frac{\pi}{2}\right) \cos^q\left(\frac{\pi}{2}\right) = 0 \]

3. At \( \theta = \tan^{-1}\left(\frac{p}{q}\right) \):
\[ f\left(\tan^{-1}\left(\frac{p}{q}\right)\right) = \sin^p\left(\tan^{-1}\left(\frac{p}{q}\right)\right) \cos^q\left(\tan^{-1}\left(\frac{p}{q}\right)\right) \]

Using the identities for sine and cosine in terms of tangent, we can express this in terms of \( p \) and \( q \).

### Step 6: Conclusion
From the evaluations, we conclude that:

- The function \( f(\theta) \) has a minimum value of 0 at both endpoints \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \).
- The maximum value occurs at \( \theta = \tan^{-1}\left(\frac{p}{q}\right) \).