11. Find the equation of the tangent line to the graph of $$ f(x)=\sin x \cos x $$ passing through the point $$ \left(\frac{3 \pi}{4},-\frac{1}{2}\right) $$.

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We start with
$$
f(x)=\sin x\cos x.
$$
Since the given point $$\left(\frac{3\pi}{4},-\frac{1}{2}\right)$$ lies on the graph (as will be verified below), the tangent line at this point is the line we seek.

1. Verify the point is on the graph. Compute:
   $$
   f\left(\frac{3\pi}{4}\right)=\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac{3\pi}{4}\right).
   $$
   We know
   $$
   \sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2} \quad \text{and} \quad \cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}.
   $$
   Hence,
   $$
   f\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{2}{4}=-\frac{1}{2}.
   $$
   This matches the given $$y$$-coordinate.

2. Compute the derivative $$f'(x)$$ to find the slope of the tangent line. It is often easier to work with an equivalent form of $$f(x)$$. Notice that
   $$
   f(x)=\sin x\cos x=\frac{1}{2}\sin(2x).
   $$
   Then the derivative is
   $$
   f'(x)=\frac{1}{2}\cdot 2\cos(2x)=\cos(2x).
   $$

3. Evaluate the derivative at the point $$x=\frac{3\pi}{4}$$:
   $$
   f'\left(\frac{3\pi}{4}\right)=\cos\Bigl(2\cdot \frac{3\pi}{4}\Bigr)=\cos\left(\frac{3\pi}{2}\right).
   $$
   Since
   $$
   \cos\left(\frac{3\pi}{2}\right)=0,
   $$
   the slope of the tangent line at this point is 0.

4. The equation of the tangent line in point-slope form is given by
   $$
   y-f\left(\frac{3\pi}{4}\right)=f'\left(\frac{3\pi}{4}\right)\left(x-\frac{3\pi}{4}\right).
   $$
   Substitute the known values:
   $$
   y-\left(-\frac{1}{2}\right)=0\cdot\left(x-\frac{3\pi}{4}\right).
   $$
   This simplifies to:
   $$
   y+\frac{1}{2}=0.
   $$

5. Finally, solve for $$y$$:
   $$
   y=-\frac{1}{2}.
   $$

Thus, the equation of the tangent line is
$$
y=-\frac{1}{2}.
$$
The equation of the tangent line is $$ y = -\frac{1}{2} $$.

Find the equation of the tangent line to the graph of passing through the point .

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