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

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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:

$$
y = 2\sqrt{2}x - \frac{3\sqrt{2}\pi}{4} + \sqrt{2}
$$
The equation of the tangent line to $$ f(x) = \csc(2x) $$ at $$ x = \frac{3\pi}{8} $$ is: $$ y = 2\sqrt{2}x - \frac{3\sqrt{2}\pi}{4} + \sqrt{2} $$

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