Find maximum and minimum values of `9cos^(2)x + 48 sinx cosx - 5 sin^(2)x - 2`.

Answer

To find the maximum and minimum values of the function \( y = 9 \cos^2 x + 48 \sin x \cos x - 5 \sin^2 x - 2 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the given function:
\[ y = 9 \cos^2 x + 48 \sin x \cos x - 5 \sin^2 x - 2 \]
We can use the identity \( \cos^2 x = 1 - \sin^2 x \) to rewrite \( \cos^2 x \):
\[ y = 9 (1 - \sin^2 x) + 48 \sin x \cos x - 5 \sin^2 x - 2 \]
Expanding this gives:
\[ y = 9 - 9 \sin^2 x + 48 \sin x \cos x - 5 \sin^2 x - 2 \]
Combining like terms:
\[ y = 7 - 14 \sin^2 x + 48 \sin x \cos x \]

### Step 2: Use trigonometric identities
We can use the double angle identities:
- \( \sin 2x = 2 \sin x \cos x \)
- \( \cos 2x = \cos^2 x - \sin^2 x \)

We rewrite \( \sin x \cos x \) as \( \frac{1}{2} \sin 2x \):
\[ y = 7 - 14 \sin^2 x + 24 \sin 2x \]

### Step 3: Substitute \( \sin^2 x \)
Using \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ y = 7 - 14 \left(\frac{1 - \cos 2x}{2}\right) + 24 \sin 2x \]
This simplifies to:
\[ y = 7 - 7 + 7 \cos 2x + 24 \sin 2x \]
Thus, we have:
\[ y = 7 \cos 2x + 24 \sin 2x \]

### Step 4: Find maximum and minimum values
The expression \( y = a \cos 2x + b \sin 2x \) can be maximized and minimized using the formula:
\[ \text{Maximum} = \sqrt{a^2 + b^2}, \quad \text{Minimum} = -\sqrt{a^2 + b^2} \]
where \( a = 7 \) and \( b = 24 \).

Calculating:
\[ \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

### Step 5: Conclusion
Thus, the maximum and minimum values of \( y \) are:
\[ \text{Maximum value} = 25, \quad \text{Minimum value} = -25 \]