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

The slope of the tangent line to a function #y# at #x=a# is found by calculating the value of #dy/dx#, the derivative of #y#, at #x=a#.

Where

#y=sinx#

the derivative is given by

#dy/dx=cosx#

The slope of the tangent line to #y# at #x=pi/6# is found by evaluating the derivative of #y# at #x=pi/6#:

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

The slope of the tangent line is #sqrt3/2#. Writing the equation of the line that passes through #(pi/6,1/2)# with slope #sqrt3/2# in point-slope form, we get:

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

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

Let the equation of the tangent line be #y=ax+b#

#dy/dx=cos x#

#a=cos pi/6=sqrt(3)/2#

#:. y=sqrt(3)/2x+b#

#1/2=sqrt(3)/2*pi/6+b#

#b=1/2-(sqrt(3)pi)/12#

Hence the equation of the tangent line is #y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12#

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, #y=ax+b#, a represents the gradient / slope of the line and b represents the y-intercept.

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, #a=dy/dx# when #x=pi/6# and the rest of the equation can be derived through algebra