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#? Calculus Derivatives Tangent Line to a Curve 1 Answer mizoo Jul 23, 2018 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 # https://www.desmos.com/calculator/su4x36cafg Answer link Related questions How do you find the equation of a tangent line to a curve? How do you find the slope of the tangent line to a curve at a point? How do you find the tangent line to the curve #y=x^3-9x# at the point where #x=1#? How do you know if a line is tangent to a curve? How do you show a line is a tangent to a curve? How do you find the Tangent line to a curve by implicit differentiation? What is the slope of a line tangent to the curve #3y^2-2x^2=1#? How does tangent slope relate to the slope of a line? What is the slope of a horizontal tangent line? How do you find the slope of a tangent line using secant lines? See all questions in Tangent Line to a Curve Impact of this question 4813 views around the world You can reuse this answer Creative Commons License