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

2 Answers
Aug 31, 2017

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

Explanation:

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)

Aug 31, 2017

Differentiate y and evaluate dy/dx at x=pi/6

The equation of the tangent line would then be y=ax+b, where a is the derivative at x=pi/6 and b can be solved by setting y=1/2, x=pi/6

The equation would be y=sqrt(3)/2x+1/2-(sqrt(3)pi)/12

Explanation:

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