# How do you find the equation of the tangent and normal line to the curve y=sinx at x=pi/3?

Dec 15, 2016

The tangent is:

$y \left(x\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \left(x - \frac{\pi}{3}\right)$

and the normal is:

$y \left(x\right) = \frac{\sqrt{3}}{2} - 2 \left(x - \frac{\pi}{3}\right)$

#### Explanation:

Given a (differentiable) curve $y = f \left(x\right)$ the equation of the tangent line in the point $\left({x}_{0} , {y}_{0}\right)$ where ${y}_{0} = f \left({x}_{0}\right)$ is given by:

$y \left(x\right) = f \left({x}_{0}\right) + f ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

and consequently the equation of the normal line is:

$y \left(x\right) = f \left({x}_{0}\right) - \frac{1}{f ' \left({x}_{0}\right)} \left(x - {x}_{0}\right)$

As:

${x}_{0} = \frac{\pi}{3}$
${y}_{0} = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$f ' \left(x\right) = \cos x$

$f ' \left({x}_{0}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

We have that the tangent is:

$y \left(x\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \left(x - \frac{\pi}{3}\right)$

and the normal is:

$y \left(x\right) = \frac{\sqrt{3}}{2} - 2 \left(x - \frac{\pi}{3}\right)$ 