# Thread: Area of Triangle #2

1. ## Area of Triangle #2

The tangent to the function $y=2sinxcosx$ at $x= pi /2$ and the x- and y-axes form a triangle. Determine the exact area of the triangle.

2. Note that
$y = 2\sin x\,\cos x = \sin 2x$

Plug in pi/2 into this equation so that I can get a point (pi/2, ...) -- I'll call this point A.

Find y' (should be simple enough), and then plug in pi/2 to get the slope of the tangent line (m) at that point.

As I said in another thread, the equation of the tangent line is
$y = m(x - x_1) + y_1$

Plug in m, x1, y1 (these are the coordinates of point A), and x = 0 to solve for y (ie. find the y-intercept). You should already notice that you don't need to find the x-intercept using this equation of the tangent line.

It should then be straightforward to find the distances between the origin and each of the two points, so you can use the area of a triangle formula again.