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

Oct 28, 2015

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

#### Explanation:

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

Substitute in the x value to find the y value.

$f \left(\frac{3 \pi}{2}\right) = {\sin}^{2} \left(\frac{3 \pi}{2}\right) = {\left(\sin \left(\frac{3 \pi}{2}\right)\right)}^{2} = {\left(- 1\right)}^{2} = 1$

The coordinates of the point are $\left(\frac{3 \pi}{2} , 1\right)$

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

Find the equation for the derivative(slope)

$f ' \left(x\right) = 2 \left(\sin x\right) \left(\cos x\right)$

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

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

$f ' \left(x\right) = 2 \sin \left(\frac{3 \pi}{2}\right) \cos \left(\frac{3 \pi}{2}\right) = 2 \left(- 1\right) \left(0\right) = - 2 \left(0\right) = 0 = m$

A slope of zero indicates a horizontal line.

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

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$ where $\left({x}_{1} , {y}_{1}\right)$ indicates the coordinates of the point.

$m = s l o p e = 0$

Now substitute in the values

$\left(y - 1\right) = 0 \left(x - \frac{3 \pi}{2}\right)$

$\left(y - 1\right) = 0$

$y - 1 = 0$

$y = 1$, 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]}