What is the slope of the line normal to the tangent line of #f(x) = xcotx-sin^2(x-pi) # at # x= (pi)/12 #?

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Answer 1 · 2018-07-23T15:48:18

#y=(3/2+sqrt(3)-2/3*pi-pi/sqrt(3))x+1/72(-36+18*sqrt(3)+3pi+4pi^2+2sqrt(3)*pi^2)#

given is

#f(x)=xcot(x)-sin^2(x-pi)# then we get by the sum , the chain and the product rule:

#f'(x)=cot(x)-xcsc^2(x)-2sin(x)cos(x)#

now we compute

#f'(pi/12)=3/2+sqrt(3)-2/3*pi-pi/sqrt(3)#

and

#f(pi/12)=1/12(2+sqrt(3))pi-1/8*(sqrt(3)-1)^2#

The Tangent line is given by the equation

#y=mx+n#

#m=f'(pi/12)#

plugging #x=pi/12# and #f(pi/12)# in the given equation we get

#n=1/72(-36+18sqrt(3)+3pi+4pi^2+2sqrt(3)pi^2)#