Simplifying Trigonometric Functions: Radians vs Degrees Explained

[Song]

In Cal, for the arctan(1/sqrt(3)), can I write it in degree or it's must be in radians?

here I have sqrt(3)*arctan(1/sqrt(3))-arctan(1)-integral (with the upper bound sqrt(3) and the lower bound 1) (x^3/(1+x^2)) dx

if I change it to degree
so I have 30sqrt(3)-45-1+(1/2)In2

But if it's in radians it would be sqrt(3)*Pi/g-Pi/4-1+(1/2)ln2

I got this pro wrong simple because I change that tan thing into degree...but what's wrong with that?

 

[HallsofIvy]

Generally speaking, when a problem involves trig functions as functions and does not involve angles or triangles, the argument is assumed to be in radians (actually, its deeper than that- technically, there are no units but radians give the right number!).