Hi,

I'm a little confused about the recitation in Session 15 (Derivative of the Arccos function). Why does Joel draw the angle arccos(x) like it was the same as cos(x)? They seem to be treated as equivalent when he draws the right-angled triangle, but I thought arccos is the inverse of cos.

Question

Hi,

I'm a little confused about the recitation in Session 15 (Derivative of the Arccos function). Why does Joel draw the angle arccos(x) like it was the same as cos(x)? They seem to be treated as equivalent when he draws the right-angled triangle, but I thought arccos is the inverse of cos.

anonymous · Answer

x is the distance of the line on the bottom of the triangle so $$cos(\theta) = x$$
x being the distance, cosine theta gives that distance
now if you plug that distance into arccos then it will return the original angle theta like so
$$\arccos(x) = \theta$$
so Joel is correct, he could have just have said theta but that is not really getting across his point.