+956 
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!!!
+129852 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
