Okay, so what I did was first plugged in π/3 into y=x-tan x and got: (π/3, -0.68)

Then I found the derivative of y and got y' = 1-sec^2x

I plugged in π/3 in y' and got -3

So using my slope -3 and my point (π/3, -0.68) I did the following:

-0.68 = -3(π/3)+b

b = 2.5

So my equation is: y = -3x+2.5

But as you can see it doesn't go through π/3

Am I doing something wrong?

Thanks for all your help!!!

Julius
Apr 16, 2018

#1**+2 **

y = x - tan x

y' = 1 - sec^2x

At pi/3 the slope is 1 - sec^2 (pi/3) = 1 - (2)^2 = 1 - 4 = -3

And...at pi/3, y = pi/3 - tan (pi/3) = pi/3 - √3

So....the equation of the tangent line at (pi/3, pi/3 - √3 ) is

y = -3 (x - pi/3) + pi/3 - √3 simplify

y = -3x + pi + pi/3 - √3

y = -3x + (4/3)pi - √3

Here's the graph, Julius : https://www.desmos.com/calculator/aheivaxqtb

CPhill
Apr 16, 2018