Thread: Trig Identities Problem

cos(arcTan (5/12) - arcCos(2/3))
so basically I set up the two triangles and get the hypotenuse 13 and the side length 2sqrt5. I'm supposed to evaluate without using a calculator, what do I do next?

First - recognize that what you have are two angles - call them alpha and beta, where:



You are aked to find . So use the trig identity:

You can figure the values of these sines and cosines from the information you've been given. For example, . BTW - the side legth of the second triangle is

Hello, hellojellojw!

Smplify: . .[1]

Let: .

We have: .

. . Hence: . .[2]


Let: .

We have: .

. . Hence: . .[3]


From [1], we have: .


Substitute [2] and [3]: .

why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?

Originally Posted by BugzLooney

why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?

Definition of the arctan: it's the angle whose tangent is the given value. Think of arctan as the inverse function of tan. Hence the tangent of the arctangent of x is simply x. You can see this with a simple example: consider the angle whose tangent = 1. You can probably see right away that the angle is . Now find the tangent of - it's 1. So you see that tan(arctan(x)) = x.