Find the equation of the tangent line to the following curves :
`x=cost, y=sint" at "t=(pi)/(4).`

Answer

To find the equation of the tangent line to the curves given by \( x = \cos t \) and \( y = \sin t \) at the point where \( t = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Find the coordinates of the point at \( t = \frac{\pi}{4} \)

1. Substitute \( t = \frac{\pi}{4} \) into the equations for \( x \) and \( y \):
\[ x = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
\[ y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, the point is \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).

### Step 2: Find the slope of the tangent line

2. Differentiate \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = -\sin t \]
\[ \frac{dy}{dt} = \cos t \]

3. The slope \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t \]

4. Substitute \( t = \frac{\pi}{4} \) into the slope:
\[ \frac{dy}{dx}\bigg|_{t=\frac{\pi}{4}} = -\cot\left(\frac{\pi}{4}\right) = -1 \]

### Step 3: Write the equation of the tangent line

5. Use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) and \( m = -1 \):
\[ y - \frac{1}{\sqrt{2}} = -1\left(x - \frac{1}{\sqrt{2}}\right) \]

6. Simplifying this equation:
\[ y - \frac{1}{\sqrt{2}} = -x + \frac{1}{\sqrt{2}} \]
\[ y + x = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]

Thus, the equation of the tangent line is:
\[ x + y = \sqrt{2} \]