The greatest value of the function `y= sin^(2) x -20cos x +1` is

Answer

To find the greatest value of the function \( y = \sin^2 x - 20 \cos x + 1 \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(\sin^2 x) - 20 \frac{d}{dx}(\cos x) + \frac{d}{dx}(1) \]

Using the chain rule, we find:

\[ \frac{dy}{dx} = 2 \sin x \cos x - 20 (-\sin x) + 0 \]

This simplifies to:

\[ \frac{dy}{dx} = 2 \sin x \cos x + 20 \sin x \]

### Step 2: Factor the derivative
We can factor out \( \sin x \):

\[ \frac{dy}{dx} = \sin x (2 \cos x + 20) \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ \sin x (2 \cos x + 20) = 0 \]

This gives us two cases to consider:

1. \( \sin x = 0 \)
2. \( 2 \cos x + 20 = 0 \)

### Step 4: Solve for \( x \)
**Case 1:** \( \sin x = 0 \)

This occurs at:

\[ x = n\pi \quad (n \in \mathbb{Z}) \]

**Case 2:** \( 2 \cos x + 20 = 0 \)

Solving for \( \cos x \):

\[ 2 \cos x = -20 \implies \cos x = -10 \]

Since the cosine function can only take values between -1 and 1, this case does not yield any valid solutions.

### Step 5: Evaluate the function at critical points
Now we evaluate the function \( y \) at the critical points from Case 1:

\[ y = \sin^2(n\pi) - 20 \cos(n\pi) + 1 \]

Since \( \sin(n\pi) = 0 \) and \( \cos(n\pi) = (-1)^n \):

\[ y = 0 - 20(-1)^n + 1 \]

This simplifies to:

\[ y = 20(-1)^n + 1 \]

### Step 6: Determine the greatest value
Now we analyze the values of \( y \):

- For even \( n \) (e.g., \( n = 0, 2, 4, \ldots \)):
\[ y = 20(1) + 1 = 21 \]

- For odd \( n \) (e.g., \( n = 1, 3, 5, \ldots \)):
\[ y = 20(-1) + 1 = -19 \]

Thus, the greatest value of \( y \) occurs when \( n \) is even, giving us:

\[ \text{Greatest value of } y = 21 \]

### Final Answer:
The greatest value of the function \( y = \sin^2 x - 20 \cos x + 1 \) is **21**.

---