Question

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

Answer 1

`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)`

Answer 2

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