Question

For `f(x)=sin^2x` what is the equation of the tangent line at `x=(3pi)/2`?

Answer 1

`y=1`, this is the equation of the tangent line

Explanation:

`f(x)=sin^2x=(sin x)^2`
graph{(y-(sinx)^2)=0 [-10, 10, -5, 5]}

Substitute in the x value to find the y value.

`f((3pi)/2)=sin^2((3pi)/2)=(sin ((3pi)/2))^2=(-1)^2=1`

The coordinates of the point are `((3pi)/2,1)`

graph{((x-4.71225)^2+(y-1)^2-0.007)=0 [-10, 10, -5, 5]}

Find the equation for the derivative(slope)

`f'(x)=2(sinx)(cos x)`

graph{2sinxcosx [-10, 10, -5, 5]}

Substitute in the value of x to get a numeric value for the derivative(slope)

`f'(x)=2sin((3pi)/2)cos ((3pi)/2)=2(-1)(0)=-2(0)=0=m`

A slope of zero indicates a horizontal line.

Use the point slope formula to find the equation of the tangent line.

`(y-y_1)=m(x-x_1)` where `(x_1,y_1)` indicates the coordinates of the point.

`m=slope=0`

Now substitute in the values

`(y-1)=0(x-(3pi)/2)`

`(y-1)=0`

`y-1=0`

`y=1`, this is the equation of the tangent line

graph{(y-(sinx)^2)(y-1)((x-4.71225)^2+(y-1)^2-0.007)=0 [-10, 10, -5, 5]}