How do you find an equation of the tangent line to the curve at the given point #y = 4 cos x# and #x=pi/6#?

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How do you find an equation of the tangent line to the curve at the given point #y = 4 cos x# and #x=pi/6#?

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Answer 1 · 2018-07-23T18:52:36

Find #y', y'(pi/6) and y(pi/6)#, then find the equation of the line passes through #(pi/6, y(pi/6)) # with the slope #m = y'(pi/6)#. Look the Explanation box below:

Explanation:

#y' = -4sin x#

#at x = pi/6: #

#y = 4cos (π/6)#
#y = 2sqrt 3#

#y' = -4sin (pi/6)#
#y' = -2 #

#(pi/6, 2sqrt 3) and m = -2#
#y - 2sqrt 3 = -2(x - pi/6)#
#y = -2x + pi/3 +2sqrt 3 #

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How do you find an equation of the tangent line to the curve at the given point #y = 4 cos x# and #x=pi/6#?

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