# How do you find the equation of the tangent line #y=sinx# at #(pi/6, 1/2)#?

##### 2 Answers

#### Explanation:

The slope of the tangent line to a function

Where

#y=sinx#

the derivative is given by

#dy/dx=cosx#

The slope of the tangent line to

#m=dy/dx|_(x=pi/6)=cos(pi/6)=sqrt3/2#

The slope of the tangent line is

#y-y_1=m(x-x_1)#

#y-1/2=sqrt3/2(x-pi/6)#

Differentiate y and evaluate

The equation of the tangent line would then be

The equation would be

#### Explanation:

Let the equation of the tangent line be

Hence the equation of the tangent line is

You can verify this answer visually too

graph{(y-sqrt(3)/2x-1/2+(sqrt(3)pi)/12)(y-sin(x))=0 [-1.259, 1.781, -0.477, 1.04]}

The reason the equation of a tangent line is as shown above is because in a linear function,

By definition, the gradient of a tangent line is equal to the slope of a curve at the point where the tangent line meets the curve.

Hence,