Question

What is the equation of the line tangent to `f(x)=cosx-sinx` at `x=pi/3`?

Answer 1

`y=(-sqrt3-1)/2x+(1-sqrt3)/2+(sqrt3+1)/2(pi/3)`

Explanation:

First, find the point of tangency. We know the `x` value we now have to find the `y` value.

Substitube in the given value of `x` to find `y`.

`f(pi/3)=cos(pi/3)-sin(pi/3)`

Based on what you know about the unit circle and these tutorials, https://www.youtube.com/playlist?list=PLsX0tNIJwRTyXncFO4Z5bxme0mBe2wq0R.

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

`f(pi/3)=1/2-sqrt3/2=(1-sqrt3)/2`

Point of Tangency `(pi/3,(1-sqrt3)/2)`

Now we need to find the equation of the slope by applying the derivative to `f(x)`.

`f'(x)=-sin(x)-cos(x)`

Let substitute in `pi/3` to get the numeric value of the slope

`f'(x)=-sin(pi/3)-cos(pi/3)`

`f'(x)=-sqrt3/2-1/2`

`f'(x)=(-sqrt3-1)/2=m`

Now we have to figure out the equation of tangent line by using the slope intercept formula, `y=mx+b`

`(1-sqrt3)/2=(-sqrt3-1)/2(pi/3)+b`

Isolate the variable, `b`

`(1-sqrt3)/2+(sqrt3+1)/2(pi/3)=b`

Equation of the tangent line, `=> y=mx+b`

`y=(-sqrt3-1)/2x+(1-sqrt3)/2+(sqrt3+1)/2(pi/3)`

The 2 images below are to show the point of tangency on the graph.

The first equation in blue is `f(x)`
The second equation in red is the tangent line.

The image below is the graph of all of this work.

Lastly, I have several tutorials on how to find the equation of the tangent line.

https://www.youtube.com/playlist?list=PLsX0tNIJwRTw2e-oelZuY6TGevXWsTFJ3