# What is the equation of the line normal to  f(x)=sin(2x+pi) at  x=pi/3?

Aug 10, 2016

Step 1: Determine the corresponding y coordinate that the function and the tangent pass through

This can be done by evaluating $x = a$ inside the function, where here $a = \frac{\pi}{3}$.

$f \left(\frac{\pi}{3}\right) = \sin \left(2 \left(\frac{\pi}{3}\right) + \pi\right)$

$f \left(\frac{\pi}{3}\right) = \sin \left(\frac{2 \pi}{3} + \pi\right)$

Evaluate using the sum and difference formulae:

$f \left(\frac{\pi}{3}\right) = \sin \left(\frac{2 \pi}{3}\right) \cos \pi + \sin \pi \cos \left(\frac{2 \pi}{3}\right)$

$f \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \times - 1 + 0 \times - \frac{1}{2}$

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

Note we could have also combined $\frac{2 \pi}{3}$ and $\pi$ and then evaluated, as well.

So, the function, the tangent and the normal pass through the point $\left(\frac{\pi}{3} , - \frac{\sqrt{3}}{2}\right)$.

Step 2: Determine the derivative

We will use the chain rule to find the derivative of this function.

Let $f \left(x\right) = \sin \left(u\right)$ and $u = 2 x + \pi$.

The derivative of $f \left(x\right)$ is $f ' \left(x\right) = \cos \left(u\right)$ and the derivative of $u$ is $u ' = 2$ (for $\pi$ is a constant, therefore its derivative is $0$).

The derivative of the entire function is then $f ' \left(x\right) = \cos \left(2 x + \pi\right) \times 2 = \textcolor{t u r q u o i s e}{2 \cos \left(2 x + \pi\right)}$.

Step 3: Determine the slope of the tangent

The slope of the tangent is given by evaluating $x = a$ inside the derivative. You probably find yourself asking at the moment: what does the the tangent have to do with the normal line? The answer: the tangent is perpendicular to the normal line. Or, in other words, the slope of the normal line is the negative reciprocal of that of the tangent.

$f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(2 \left(\frac{\pi}{3}\right) + \pi\right)$

This time I will use the combination inside the parentheses method.

$f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(\frac{2 \pi}{3} + \frac{3 \pi}{3}\right)$

$f ' \left(\frac{\pi}{3}\right) = 2 \cos \left(\frac{5 \pi}{3}\right)$

$f ' \left(\frac{\pi}{3}\right) = 2 \times \frac{1}{2}$

$f ' \left(\frac{\pi}{3}\right) = 1$

Hence, the slope of the tangent is $1$. The negative reciprocal of $1$ is $- \frac{1}{1} = - 1$.

$\therefore$ The slope of the normal line is of $- 1$.

Step 4: Use point-slope form to determine the equation of the normal line

We know the point that the function, tangent and normal pass through, and the slope of normal line. We also know that the normal is a line or a linear function. Hence, we can use point-slope form to determine the normal line's equation.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

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

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

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

$y = - x + \frac{2 \pi - 3 \sqrt{3}}{6}$

If you insist on an approximation, the equation of the line is $y = - x + 0.18$, but normally the exact value answer is better.

Since this is my ${900}^{\text{th}}$ answer for Socratic, I have decided to include extra practice exercises. Make good use of them! You'll improve exponentially!

Practice exercises:

1. Determine the equations of the tangents to the following relations at the given points:

a) $y = \sqrt{{x}^{3} + 9}$, at $x = - 2$

b) $y = {2}^{x - 3}$ at $x = 5$

c) $y = {x}^{5} + 3 {x}^{4} - 2 {x}^{3} - 8 {x}^{2} + 2 x - 1$ at $x = 1$

d) $y = \sin \left(2 x\right)$ at $x = \frac{\pi}{4}$

$2$. Determine the equations of the normal lines to the following relations at the given points:

a) $y = \sqrt{\frac{1}{3} {x}^{3} + 2 x + 1}$ at the point $x = 3$

b) $y = {\log}_{3} \left(2 x + 11\right)$ at the point $x = 8$

c) $y = \tan \left(4 x + \frac{\pi}{2}\right)$ at the point $\frac{\pi}{8}$

d) ${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$ at the point $\left(6 , 1\right)$

Hopefully this helps, and good luck! 