How do you find an equation of the tangent line to the curve at the given point # y=5cos(x)# and #x=pi/3#?

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Answer 1 · 2018-01-03T08:47:49

#y=-x((5sqrt(3))/2)+((5pisqrt(3)+15)/6)#

First we need to find the gradient at #x=pi/3#, since this will be the gradient of the tangent line. We need to calculate the first derivative.

#:.#

#d/dx(5cos(x))=-5sin(x)#

Gradient at #x=pi/3#

#f'(pi/3)=-5sin(pi/3)#

Slope point equation of a line.

#y-5cos(pi/3)=-5sin(pi/3)(x-pi/3)#

#y=-5xsin(pi/3)+((5pi)/3)sin(pi/3)+5cos(pi/3)#

Using identities:

#color(red)(sin(A-B)=sinAcosB-cosAsinB)#

#color(red)(cos(A-B)=cosAcosB+sinAsinB)#

#sin(pi/3)=sin(pi-(2pi)/3)=#

#->sin(pi)cos((2pi)/3)-cos(pi)sin((2pi)/3)=#

#->(0)(-1/2)-((-1)(sqrt(3)/2))=(sqrt(3))/2#

#cos(pi/3)=cos(pi-(2pi)/3)=#

#->cos(pi)cos((2pi)/3)+sin(pi)sin((2pi)/3)#

#->(-1)(-1/2)+(0)(sqrt(3)/2)=1/2#

#:.#

#y=-5xsin(pi/3)+((5pi)/3)sin(pi/3)+5cos(pi/3)#

#y=-5x((sqrt(3))/2)+((5pi)/3)((sqrt(3))/2)+5/2#

#color(blue)(y=-x((5sqrt(3))/2)+((5pisqrt(3)+15)/6))#

GRAPH:

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