Calculating inverse trig expressions like cos(arctan -2)

Question

I have some problems "connecting dots". All feedback is welcomed and really, really helpful! 🙂

Task 1: calculate $\quad \tan{(\arcsin{(-\frac{3}{4}}))}$

Solution:

$\tan{(\arcsin{-\frac{3}{4}})} = \tan{(\arctan{(- \beta)})}$

$\arctan{(- \beta)} = \quad ?$

Drawing the right triangle:

$3^2 + x^2 = 4^2$

$x = \sqrt{7} \quad$ (neglecting the negative one because side of a triangle has to be positive, right?)

$\Rightarrow \beta = \frac{3}{\sqrt{7}}$

$\Longrightarrow \tan{(\arcsin{-\frac{3}{4}})} = \tan{(\arctan{(- \beta)})} = \tan{(\arctan{(- \frac{3}{\sqrt{7}})})} = – \frac{3}{\sqrt{7}}$

And this is a good answer. But here's another example, same method, wrong answer.

Task 2: calculate $\quad \cos{(\arctan{(-2)})}$

$\cos{(\arctan{(-2)})} = \cos{(\arccos{(- \beta)})}$

$\arccos{(- \beta)} = \quad$ ?

Drawing the right triangle:

$2^2 + 1^2 = x^2$

$x = \sqrt{5}$

$\Rightarrow \beta = \frac{1}{\sqrt{5}}$

$\Longrightarrow \cos{(\arctan{(-2)})} = \cos{(\arccos{(-\beta)})} = \cos{(\arccos{(-\frac{1}{\sqrt{5}})})} = -\frac{1}{\sqrt{5}} $

And this is a bad answer… Good one should be $\frac{1}{\sqrt{5}}$ (without the minus sign). Where did I make a mistake? What is bad with my method? Can you introduce me to another method?

I am still super new to mathematics, thus the stupid mistakes. Thanks.

Answer 1

Let $x= \arctan(-2)$ then $ \tan x = -2$ and $x\in(-\pi/2,\pi/2)$

You are searching for $\cos x$. Remember that we have $${1\over \cos^2x } =1+\tan^2x \implies...$$