Question

Evaluate cos(tan^-1 0)

Answer 1

\[\large \cos(\arctan 0)\]

Hmm ok so let's first look at the inner part, \(\arctan 0\).

Since it's an inverse trig function, the solution is actually an angle. We can write it like this.\[\large \arctan 0 = \theta\]

So our problem is actually,\[\large \cos(\arctan 0) \qquad \rightarrow \qquad \cos(\theta)\]
So first of all what is the arctan of zero?
If you're not sure, you could rewrite as tangent undoing the inverse.
\[\large \arctan 0 =\theta \qquad \rightarrow \qquad \tan \theta= 0\]This occurs when theta is zero.

So our problem is now,\[\large \cos(\arctan 0) \qquad \rightarrow \qquad \cos(0)\]Remember your special angle for this one? :)

Answer 2

So:\[\cos (\tan^{-1} (0))\]
You probably know: tan 0 =0.
That gives you tan^-1(0) as well.

Answer 3

`right

Answer 4

So tan^-1(0)=0 and you are to find cos(0)=...?

Answer 5

1

Answer 6

Done!