Question

What is the equation of the tangent line of `f(x)=cosxsinx` at `x=pi/3`?

Answer 1

`y=-1/2x+pi/6+sqrt3/4`

Explanation:

`f(pi/3) = cos(pi/3)sin(pi/3) = (1/2)(sqrt3/2) = sqrt3/4`.

So the line we seek contains the point `(pi/3, sqrt3/4)`.

The slope is given by the derivative. We could differentiate by using the chain rule, but let's use some trigonometry to rewrite the function instead.

`sin(2theta) = 2sinthetacostheta`,

so `f(x) = cosxsinx=1/2sin(2x)`

And `f'(x) = cos(2x)`.

At `x=pi/3` we get slope = `cos((2pi)/3)= -1/2`

The line through `(pi/3, sqrt3/4)`, with slope `m= -1/2` is

`y=-1/2x+pi/6+sqrt3/4`.