2

The problem:

Simplify the expression. Specify the range of $x$ for which the simplification holds: $\cos(\tan^{-1}x)$.

So we know that, $\tan^{-1}x$ is the angle $\theta$ for which $\tan\theta = x$.

So I sketched a triangle, much like I would as if it were in the unit circle.

I have one side, the one from the origin that would touch the unit circle, which has the length of 1.

I think that the x-axis is $\cos\theta$ and the y-axis is $\sin\theta$. Is that correct? I'm stuck here, not sure what to do next.

Martin Sleziak
  • 53,687
user2451412
  • 117
  • 1
  • 10

2 Answers2

1

Draw a triangle with adjacent side $1$ and opposite side $x$:

$\theta=\tan^{-1}x\Leftrightarrow\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{1}$.

Now, since we know both the opposite and adjacent side, we can now compute the hypotenuse by the Pythagorean Theorem: $\text{hypothenuse}=\sqrt{x^2+1}$.

$\cos(\tan^{-1}x)=\cos\theta=\frac{\text{adjacent side}}{\text{hypothenuse}}=\frac{1}{\sqrt{x^2+1}}$

All values of $x$ are allowed, so the domain is: $\boxed{x\in\mathbb{R}}$.

rae306
  • 9,742
0

If $\displaystyle\arctan x=\theta,\tan\theta=x$

So, $$\cos^2\theta=\frac1{\sec^2\theta}=\frac1{1+\tan^2\theta}=\cdots$$

Using the definition of principal value of trigonometric functions, $$-\frac\pi2\le\theta\le\frac\pi2\implies\cos\theta\ge0$$

lab bhattacharjee
  • 274,582