Question Given \( f(x)=\csc (2 x) \), write the equation of the line tangent to \( y=f(x) \) when \( x=\frac{3 \pi}{8} \)

To find the equation of the tangent line to the function \( f(x) = \csc(2x) \) at the point where \( x = \frac{3\pi}{8} \), we need to follow these steps: 1. **Calculate \( f\left(\frac{3\pi}{8}\right) \)** to find the y-coordinate of the point of tangency. 2. **Find the derivative \( f'(x) \)** to determine the slope of the tangent line at that point. 3. **Evaluate \( f'\left(\frac{3\pi}{8}\right) \)** to get the slope at \( x = \frac{3\pi}{8} \). 4. **Use the point-slope form of the equation of a line** to write the equation of the tangent line. Let's start with step 1. ### Step 1: Calculate \( f\left(\frac{3\pi}{8}\right) \) \[ f\left(\frac{3\pi}{8}\right) = \csc\left(2 \cdot \frac{3\pi}{8}\right) = \csc\left(\frac{3\pi}{4}\right) \] The cosecant function is the reciprocal of the sine function: \[ \csc\left(\frac{3\pi}{4}\right) = \frac{1}{\sin\left(\frac{3\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \] So, we have: \[ f\left(\frac{3\pi}{8}\right) = \sqrt{2} \] ### Step 2: Find the derivative \( f'(x) \) To find the derivative of \( f(x) = \csc(2x) \), we use the chain rule. The derivative of \( \csc(u) \) is \( -\csc(u) \cot(u) \cdot u' \), where \( u = 2x \) and \( u' = 2 \). Thus, \[ f'(x) = -\csc(2x) \cot(2x) \cdot 2 = -2 \csc(2x) \cot(2x) \] ### Step 3: Evaluate \( f'\left(\frac{3\pi}{8}\right) \) Now we need to evaluate the derivative at \( x = \frac{3\pi}{8} \): \[ f'\left(\frac{3\pi}{8}\right) = -2 \csc\left(2 \cdot \frac{3\pi}{8}\right) \cot\left(2 \cdot \frac{3\pi}{8}\right) = -2 \csc\left(\frac{3\pi}{4}\right) \cot\left(\frac{3\pi}{4}\right) \] We already know that \( \csc\left(\frac{3\pi}{4}\right) = \sqrt{2} \). Now we calculate \( \cot\left(\frac{3\pi}{4}\right) \): \[ \cot\left(\frac{3\pi}{4}\right) = \frac{\cos\left(\frac{3\pi}{4}\right)}{\sin\left(\frac{3\pi}{4}\right)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 \] Now substituting these values back into the derivative: \[ f'\left(\frac{3\pi}{8}\right) = -2 \cdot \sqrt{2} \cdot (-1) = 2\sqrt{2} \] ### Step 4: Write the equation of the tangent line We have the point \( \left(\frac{3\pi}{8}, \sqrt{2}\right) \) and the slope \( 2\sqrt{2} \). Using the point-slope form of the equation of a line: \[ y - y_1 = m(x - x_1) \] Substituting \( m = 2\sqrt{2} \), \( x_1 = \frac{3\pi}{8} \), and \( y_1 = \sqrt{2} \): \[ y - \sqrt{2} = 2\sqrt{2}\left(x - \frac{3\pi}{8}\right) \] This is the equation of the tangent line. To express it in slope-intercept form, we can rearrange it: \[ y = 2\sqrt{2}x - 2\sqrt{2} \cdot \frac{3\pi}{8} + \sqrt{2} \] Thus, the final equation of the tangent line is: \[ y = 2\sqrt{2}x - \frac{3\sqrt{2}\pi}{4} + \sqrt{2} \]