## Calculus 8th Edition

$$\boxed{y-1/2 = \sqrt{3}/2*(x-\pi/6)}$$
To find the linearization we first need to calculate the slope by taking the derivative. $$f'(\pi/6) = cos(\pi/6) = \sqrt {3}/2$$ We also need to find the y-value at the given $a$ value, $$f(\pi/6) = sin(\pi/6) = 1/2$$ Since we have the point $(\pi/6, 1./2)$ and the slope $\sqrt {3}/2$, we can use point slope form. $$\boxed{y-1/2 = \sqrt{3}/2*(x-\pi/6)}$$