# What is the equation of the line that is normal to #f(x)=-sinx +tanx # at # x=-pi/3#?

##### 1 Answer

#### Explanation:

First, find the point the normal line will intercept:

#f(-pi/3)=-sin(-pi/3)+tan(-pi/3)=sqrt3/2-sqrt3=ul(-sqrt3/2#

The normal line passes through the point

To find the slope of the normal line, we must first find the slope of the tangent line. Since the normal line and tangent line are perpendicular, their slopes will be opposite reciprocals.

To find the slope of the tangent line, find the value of the derivative at

The derivative of the function is

#f'(x)=-cosx+sec^2x#

The slope of the tangent line is

#f'(-pi/3)=-cos(-pi/3)+sec^2(-pi/3)=-1/2+2^2=ul(7/2#

Since the slopes of the tangent and normal lines are opposite reciprocals, the slope of the normal line is

The point of

#y+sqrt3/2=-2/7(x+pi/3)#

Graphed are the original function and its normal line:

graph{(y+sinx-tanx)(y+sqrt3/2+2/7(x+pi/3))=0 [-4.858, 2.94, -2.89, 1.007]}