A function and its inverse undo each other, but with inverse trig functions that is only half true, because of the restricted ranges. This lesson sorts out which way works and how to handle the other.
Idea clave
$\sin(\arcsin x) = x$ for every $x$ in $[-1, 1]$. But $\arcsin(\sin\theta) = \theta$ only when $\theta$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; otherwise it is the angle in that range with the same sine. The same goes for cosine (range $[0, \pi]$) and tangent. For expressions like $\cos(\arctan x)$, name the angle and draw a right triangle.
Por qué funciona
$\arcsin x$ is an angle whose sine is $x$, so taking its sine gives back $x$. Going the other way, $\sin\theta$ forgets which angle it came from: $\sin\frac{5\pi}{6}$ and $\sin\frac{\pi}{6}$ are both $\frac{1}{2}$. Arcsine can only answer with the angle in its range.
Worked examples
Ejemplo resuelto: Outside the range
Find $\arcsin\left(\sin\frac{5\pi}{6}\right)$.
- Inside first: $\sin\frac{5\pi}{6} = \frac{1}{2}$.
- $\arcsin\frac{1}{2}$ must be in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: it is $\frac{\pi}{6}$.
Respuesta: $\frac{\pi}{6}$, not $\frac{5\pi}{6}$
Ejemplo resuelto: A triangle
Find $\cos\left(\arctan\frac{3}{4}\right)$.
- Let $\theta = \arctan\frac{3}{4}$, so $\tan\theta = \frac{3}{4}$ and $\theta$ is in Quadrant I.
- Draw a right triangle with opposite side $3$ and adjacent side $4$; the hypotenuse is $5$.
- $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$.
Respuesta: $\frac{4}{5}$
Ejemplo resuelto: Watch the quadrant
Find $\sin\left(\arccos\left(-\frac{5}{13}\right)\right)$.
- Let $\theta = \arccos\left(-\frac{5}{13}\right)$. It is in $[0, \pi]$ with a negative cosine, so it is in Quadrant II.
- Reference triangle: adjacent $5$, hypotenuse $13$, opposite $12$.
- Sine is positive in Quadrant II: $\sin\theta = \frac{12}{13}$.
Respuesta: $\frac{12}{13}$
Ejemplo resuelto: With a variable
Write $\cos(\arctan x)$ without trig functions.
- Let $\theta = \arctan x$, so $\tan\theta = \frac{x}{1}$: opposite $x$, adjacent $1$, hypotenuse $\sqrt{1 + x^2}$.
- $\cos\theta = \frac{1}{\sqrt{1 + x^2}}$. On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ cosine is positive, so this works for negative $x$ too.
Respuesta: $\frac{1}{\sqrt{1 + x^2}}$
Error común
Cancelling $\arcsin(\sin\theta)$ to $\theta$ every time. Check the answer against the range: an arcsine can never be $\frac{5\pi}{6}$, and an arccosine can never be negative.