Find the range of `12sintheta-9sin^2theta`

Explanation

To find the range of the function \( f(\theta) = 12 \sin \theta - 9 \sin^2 \theta \), we can follow these steps:

### Step 1: Rewrite the function
Let \( y = f(\theta) = 12 \sin \theta - 9 \sin^2 \theta \).

### Step 2: Substitute \( x = \sin \theta \)
Since \( \sin \theta \) can take values between -1 and 1, we can substitute \( x \) for \( \sin \theta \):
\[ y = 12x - 9x^2 \]

### Step 3: Rearrange the expression
Rearranging gives us:
\[ y = -9x^2 + 12x \]

### Step 4: Complete the square
To find the maximum and minimum values, we can complete the square:
\[ y = -9(x^2 - \frac{12}{9}x) \]
\[ = -9\left(x^2 - \frac{4}{3}x\right) \]
Now we complete the square inside the parentheses:
\[ = -9\left(x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\right) \]
\[ = -9\left(\left(x - \frac{2}{3}\right)^2 - \frac{4}{9}\right) \]
\[ = -9\left(x - \frac{2}{3}\right)^2 + 4 \]

### Step 5: Identify the range
The term \(-9\left(x - \frac{2}{3}\right)^2\) is always less than or equal to 0, and its maximum value occurs when \( x = \frac{2}{3} \):
- When \( x = \frac{2}{3} \), \( y = 4 \).
- When \( x = -1 \), \( y = 12(-1) - 9(-1)^2 = -12 - 9 = -21 \).
- When \( x = 1 \), \( y = 12(1) - 9(1)^2 = 12 - 9 = 3 \).

Thus, the maximum value of \( y \) is \( 4 \) and the minimum value is \( -21 \).

### Conclusion
The range of the function \( f(\theta) = 12 \sin \theta - 9 \sin^2 \theta \) is:
\[ \text{Range} = [-21, 4] \]

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