What is the equation of the line tangent to # f(x)=secx - 2cosx # at # x=pi/3#?

1 Answer
Psykolord1989 .
Nov 25, 2017

Calculate the derivative (slope of the line), and find #f(pi/3)# so you have a point for point slope form to find the line equation.

#y = (3sqrt3)x + 1 - (sqrt3)pi#

Explanation:

To find the equation of the line in question, we need both a point, and a slope. The point will be found by simply finding #f(pi/3)#

#f(pi/3) = sec (pi/3) - 2 cos(pi/3) = 1/cos(pi/3)-2cos(pi/3)#

We know, either from use of a calculator or from learning the values of the trig functions for angles such as #npi, npi/2, npi/3, npi/4, npi/6#, that #cos(pi/3) = 1/2#. Thus...

#f(pi/3) = 1/(1/2) - 2(1/2) = 2-1 = 1#

Thus the point #(pi/3, 1)# is where our tangent line touches.

For the slope, we must calculate the derivative of the function, and then find the exact derivative at #pi/3#

#(df)/dx = secxtanx + 2sinx#

If #x=pi/3...#

#(df)/dx (x=pi/3) = sec (pi/3) tan (pi/3) + 2 sin (pi/3) = 1/(1/2) * ((sqrt3)/2)/(1/2) + 2(sqrt3)/2 = 2*sqrt3 + sqrt3 = 3sqrt3 #

Thus, our tangent line is of the form #y=(3sqrt3) x + c_1#

Since #y(pi/3)=1...#, use point slope form...

#y-1 = 3sqrt3 (x-pi/3) = 3sqrt3x - pisqrt3 -> y = 3sqrt3x + 1 - sqrt3pi#

Thus, the tangent line equation is:

#y = (3sqrt3)x + 1 - (sqrt3)pi#

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