# Find the equation of the tangent and normal to the curve y=cosx at x=π/3?

Jun 22, 2018

Tangent's equation is $y = - \frac{\sqrt{3}}{2} x + \frac{\pi}{2 \sqrt{3}} + \frac{1}{2}$

Normal's equation is $y = \frac{2}{\sqrt{3}} x - \frac{2 \pi}{3 \sqrt{3}} + \frac{1}{2}$

#### Explanation:

First, you find the first derivative of the function then substitute with the given point $\left({x}_{1} , {y}_{1}\right)$ to get the slope$\left(m\right)$ and then substitute in this function

color(green)((y-y_1)=m(x-x_1)

apply for the given function

$y = \cos x$

At $x = \frac{\pi}{3}$ $\rightarrow$ $y = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

$\left({x}_{1} , {y}_{1}\right) =$$\left(\frac{\pi}{3} , \frac{1}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin x$

Substitute with the point $\left(\frac{\pi}{3} , \frac{1}{2}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {m}_{1} = - \sin \left(\frac{\pi}{3}\right) = - \frac{\sqrt{3}}{2}$

Now Substitute in the formula in green with $\left({x}_{1} , {y}_{1}\right) , {m}_{1}$
to get the equation of the tangent to the curve at the given point

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

$y = - \frac{\sqrt{3}}{2} x + \frac{\pi}{2 \sqrt{3}} + \frac{1}{2}$

the normal to the curve would be at the same point but its slope is different and in order to get its slope ${m}_{2}$

We use this

${m}_{1} \cdot {m}_{2} = - 1$

${m}_{2} = \frac{2}{\sqrt{3}}$

Now Substitute with $\left({x}_{1} , {y}_{1}\right) , {m}_{2}$ in the equation in green to find the equation of the normal

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

$y = \frac{2}{\sqrt{3}} x - \frac{2 \pi}{3 \sqrt{3}} + \frac{1}{2}$