For #f(x)=sinx# what is the equation of the tangent line at #x=(3pi)/2#?

1 Answer
Dec 19, 2015

#y = -1#

Explanation:

The equation of the tangent line of any function at #x = a# is given by the formula : #y = f'(a)(x-a) + f(a)#. So we need the derivative of #f#.

#f'(x) = cos(x)# and #cos((3pi)/2) = 0# so we know that the tangent line at #x = 3pi/2# is horizontal and is #y = sin((3pi)/2) = -1#