Question

How do you find the exact value of `cos[arctan(-8/15)]`?

Answer 1

Ok guy i want you to imagine a triangle rectangle with one side `a` and another side `b`, so the hypotenuse `c` is equal to `sqrt(a^2+b^2)`

You have an angle `theta` is opposite side is `a` and his adjacent side is `b`

So you know trigonometry and

`tan(theta) = a/b`

`-tan(theta) = -a/b`

and you know `tan(theta)` is odd so

`tan(-theta) = -a/b`

So you can say

`-theta = arctan(-a/b)`

again, you know the trigonometry and know

`cos(theta) = b/c`

but you know `-theta = arctan(-a/b)` and the cosinus is an even function so you don't care if you put `-theta` or `theta` so

`cos(arctan(-a/b)) = b/c`