How do you find the equation of the tangent and normal line to the curve #y=sinx# at #x=pi/3#?

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Answer 1 · 2016-12-15T17:37:09

The tangent is:

#y(x) = sqrt(3)/2+1/2(x-pi/3)#

and the normal is:

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

Given a (differentiable) curve #y=f(x)# the equation of the tangent line in the point #(x_0, y_0)# where #y_0=f(x_0)# is given by:

#y(x) = f(x_0)+f'(x_0)(x-x_0)#

and consequently the equation of the normal line is:

#y(x) = f(x_0)-1/(f'(x_0))(x-x_0)#

As:

#x_0 = pi/3#
#y_0 =sin(pi/3) = sqrt(3)/2#

#f'(x) = cosx#

#f'(x_0) = cos(pi/3)= 1/2#

We have that the tangent is:

#y(x) = sqrt(3)/2+1/2(x-pi/3)#

and the normal is:

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

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