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

Here, it is necessary to know the domain of trig functions and their inverses. You’ve ignored them, which is exactly why you made that error. The domain of $\tan (x)$ is $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so the range of $\arctan(x)$ is $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Generally, when the tangent of an angle is negative, it can either be in quadrants $2$ or $4$, but by how the domain has been chosen for $\tan(x)$, we consider only the quadrant $4$ angle. So $\arctan(-2)$ returns some angle in quadrant $4$, and cosine is positive there. You chose the quadrant $2$ angle, where cosine is negative.

Addition: Here are the domain and range of the six main trig functions:

$$\begin{array}{|c|} \hline \text{Function}&\text{Domain}&\text{Range}\\\hline \sin(x)&-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}&\vert y\vert \leq 1\\\hline \cos(x)&0 \leq x \leq \pi&\vert y\vert \leq 1&\\ \hline \tan(x)&-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}&y \in \mathbb{R}\\\hline \csc(x)&-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}&\vert y\vert \geq 1\\\hline \sec(x)&0 \leq x \leq \pi&\vert y\vert \geq 1 \\ \hline \cot(x)&0 < x < \pi&y \in \mathbb{R}\\\hline \end{array}$$

For the inverse of each, simply switch the domain and range.

Answer 2

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 $${1\over \cos^2x } =1+\tan^2x \implies...$$

Answer 3

This a common mistake among students. As KM101 pointed out, you cannot work with right-angled triangles in general, to solve problems related to trigonometric and inverse trigonometric functions. Sine, cosine, and the forth, correspond to ratio between triangle side lenghts only when dealing with angles in the range $\left[0, \frac{\pi}{2}\right]$.

In any other case you must stick with the definitions, based, for example, on the coordinates of points belonging to the circle of radius $1$ centered at the origin of the axes.

Morover, since trigonometric functions are not invertible over their entire domain, they are "limited" conventionally to some intervals before inversion.

The sine function is limited to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, where it is monotonically increasing. Consequently $\arcsin(\cdot )$ has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Similarly, to obtain monotonicity, cosine is limited between $0$ and $\pi$, and that is going to be the range of $\arccos(\cdot )$. Finally the tangent is taken in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is going to be the range of $\arctan(\cdot )$.

So, for example, if you have to calculate $$\cos(\arctan (-2))$$ you must ask yourself what is the cosine of the angle that has tangent equal to $-2$, in the range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.This angle being in the fourth quadrant, it must have negative sine and positive cosine. Only now can you use the quadratic relationship between sine and cosine (together with the known ratio between them, i.e. $-2$) to get the correct result.

Answer 4

\begin{array}{|c|c|cc|} \hline \text{function} & \text{domain} & \text{range} & \text{quadrants} \\ \hline \sin^{-1} & [-1, 1] & [-\frac{\pi}{2}, \frac{\pi}{2}] & \text{I, IV}\\ \cos^{-1} & [-1, 1] & [0, \pi] & \text{I, II}\\ \tan^{-1} & [-\infty, \infty] & [-\frac{\pi}{2}, \frac{\pi}{2}] & \text{I, IV} \\ \cot^{-1} & [-\infty, \infty] & (0, \pi) & \text{I, II}\\ \sec^{-1} & [-\infty, -1] \cup [1, \infty] & [0, \frac{\pi}{2}] \cup [\frac{\pi}{2}, \pi] & \text{I, II} \\ \csc^{-1} & [-\infty, -1] \cup [1, \infty] & [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] & \text{I, IV} \\ \hline \end{array}

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$$\cos \arctan -2$$

It's better to use coordinates instead of triangles.

Since tangent is $\dfrac yx$, then $\dfrac yx = -2$. This puts the point $(x,y)$ in the second or fourth quadrant.

The range of arctan is in the first and fourth quadrants. So we need $(x,y) = (1, -2)$ and $r = \sqrt 5$.

So $\cos \arctan -2 = \dfrac xr = \dfrac{1}{\sqrt 5}$