Question

given the equation y= 3sin ^2 (x/2) what is an equation of the tangent line to the graph x=pi .
( I've done these in the past but I can't remember where to start, can someone just help me out with the steps)

Answer 1

Find the y coordinate of the point.

Answer 2

Then find the slope of the tangent which is the value of the first derivative when x is replaced with pi

Answer 3

After you know the slope and the point, you can write the equation of the tangent line.

Answer 4

First derivative is:\[y'=\frac{3\sin (x)}{2}\]

Answer 5

let \[y = (\sin(x/2)^2\] let u = sin(x/2) du/dx = 1/2cos(x/2)

dy/du = 2u

dy/dx = dy/du x dx/du

dy/dx = 2sin(x/2) x 1/2cos(x/2)

dy/dx = sin(x/2)cos(x/2)

sub x = pi to get the gradient...

Answer 6

When x = pi\[y'=slope=\frac{3\sin \pi}{2}=0\]

Answer 7

oops forgot the 3 at the front

Answer 8

So we see that the tangent is horizontal at the given point and its equation is y =3