Find the equation of the tangent line to the curve `y=sinx" at "x=(pi)/(4).`

Answer

To find the equation of the tangent line to the curve \( y = \sin x \) at \( x = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( y = \sin x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \cos x \]

**Hint:** The derivative of \( \sin x \) is \( \cos x \).

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{4} \)
Next, we will find the slope of the tangent line by substituting \( x = \frac{\pi}{4} \) into the derivative.

\[ \text{slope} = \frac{dy}{dx} \bigg|_{x = \frac{\pi}{4}} = \cos\left(\frac{\pi}{4}\right) \]

Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), we have:

\[ \text{slope} = \frac{1}{\sqrt{2}} \]

**Hint:** Remember that \( \cos\left(\frac{\pi}{4}\right) \) is a standard trigonometric value.

### Step 3: Find the y-coordinate at \( x = \frac{\pi}{4} \)
Now, we need to find the y-coordinate of the point on the curve at \( x = \frac{\pi}{4} \).

\[ y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

**Hint:** The sine of \( \frac{\pi}{4} \) is also \( \frac{1}{\sqrt{2}} \).

### Step 4: Use the point-slope form to write the equation of the tangent line
Now that we have the slope and the point \( \left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right) \), we can use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = \frac{1}{\sqrt{2}} \), \( x_1 = \frac{\pi}{4} \), and \( y_1 = \frac{1}{\sqrt{2}} \):

\[ y - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\left(x - \frac{\pi}{4}\right) \]

### Step 5: Simplify the equation
Now, we can simplify the equation:

\[ y - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}x - \frac{\pi}{4\sqrt{2}} \]

Adding \( \frac{1}{\sqrt{2}} \) to both sides:

\[ y = \frac{1}{\sqrt{2}}x - \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}} \]

Thus, the equation of the tangent line is:

\[ y = \frac{1}{\sqrt{2}}x + \left(1 - \frac{\pi}{4}\right)\frac{1}{\sqrt{2}} \]

**Final Answer:** The equation of the tangent line to the curve \( y = \sin x \) at \( x = \frac{\pi}{4} \) is:

\[ y = \frac{1}{\sqrt{2}}x + \left(1 - \frac{\pi}{4}\right)\frac{1}{\sqrt{2}} \]