# What is the equation of the normal line of #f(x)= cscx# at #x = pi/8#?

##### 1 Answer

Note that

The slope of the normal line will be the inverse reciprocal of the slope of the tangent line at

To find the slope of the tangent line there, find the derivative of

The derivative of

#f(x)=(sin(x))^-1#

#=>f'(x)=-(sin(x))^-2*cos(x)=-cos(x)/sin^2(x)#

So the slope of the tangent line is

The line passing through

#y-1/sin(pi/8)=sin^2(pi/8)/cos(pi/8)(x-pi/8)#

We could find these decimals, but we can also find the exact values.

I like to use the double angle formulas:

#cos(2x)=2cos^2(x)-1#

So:

#cos(pi/4)=2cos^2(pi/8)-1#

#cos^2(pi/8)=1/2(cos(pi/4)+1)=1/2(1/sqrt2+1)=(1+sqrt2)/(2sqrt2)=(2+sqrt2)/4#

Then:

#cos(pi/8)=sqrt(2+sqrt2)/2#

The same process can be done using

#sin^2(pi/8)=(2-sqrt2)/4#

#sin(pi/8)=sqrt(2-sqrt2)/2#

Then the normal line becomes:

#y-2/sqrt(2-sqrt2)=(2-sqrt2)/4(2/sqrt(2+sqrt2))(x-pi/8)#

#y-2/sqrt(2-sqrt2)=(2-sqrt2)/(2sqrt(2+sqrt2))(x-pi/8)#