## Calculus (3rd Edition)

$$y=\left(x-\frac{\pi}{4}\right)+\ln(1/\sqrt{2})$$
Given $$y=\ln (\sin x), \quad x=\frac{\pi}{4}$$ Since at $x=\pi/4$, $f(x) = \ln (1/\sqrt{2})$ and $$f'(x)= \frac{\cos x}{\sin x}= \cot x$$ Then $m= f'(x)\bigg|_{x=\pi/4}=1$ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-\ln (1/\sqrt{2}) }{x- \pi/4}&=1 \\ y&=\left(x-\frac{\pi}{4}\right)+\ln(1/\sqrt{2}) \end{align*}