Introduction
Trigonometric functions are not bijective on the real numbers, so they do not have inverse functions when mapping there. However, they can be bijective when restricted to a certain domain.
For example, here is the graph of
y=\sin(x)
, limited to the domain of
[-\frac{\pi}{2}, \frac{\pi}{2}]
, denoted by the blue lines.
Here is the same graph, using the blue lines as the cutoff.
That is bijective, mapping to the range of
[-1,1]
.
All trigonometric functions can become bijective by limiting their domain.
Their inverse functions are frequently used, so they have their own names and notations. The inverse of sine is arcsine. Cosine's is arccosine. Tangent's is arctangent, etc.
Here is the notation for the inverse trig functions and their domains. Note that there are many domains which could allow for these functions to be bijective, but these are the most common. They all include the origin.
\begin{align*}
f(x) &= \sin(x)& f^{-1}(x) &= \arcsin(x) &&\text{Domain: } [-\frac{\pi}{2}, \frac{\pi}{2}] &&\text{Range: } [-1,1] \\
f(x) &= \cos(x)& f^{-1}(x) &= \arccos(x) &&\text{Domain: } [0, \pi] &&\text{Range: } [-1,1] \\
f(x) &= \tan(x)& f^{-1}(x) &= \arctan(x) &&\text{Domain: } [-\frac{\pi}{2}, \frac{\pi}{2}] &&\text{Range: } (-\infty,\infty) \\
f(x) &= \csc(x)& f^{-1}(x) &= \text{arccsc}^{-1}(x) &&\text{Domain: } [-\frac{\pi}{2}, \frac{\pi}{2}] &&\text{Range: } (-\infty,-1] \cup [1,\infty) \\
f(x) &= \sec(x)& f^{-1}(x) &= \text{arcsec}^{-1}(x) &&\text{Domain: } [0, \pi] &&\text{Range: } (-\infty,-1] \cup [1,\infty) \\
f(x) &= \cot(x)& f^{-1}(x) &= \text{arccot}^{-1}(x) &&\text{Domain: } [0, \pi] &&\text{Range: } (-\infty,\infty)
\end{align*}
Note that
\arcsin
,
\arccos
,
\arctan
,
\text{arccsc}
,
\text{arcsec}
, and
\text{arccot}
can be written as
\sin^{-1}
,
\cos^{-1}
,
\tan^{-1}
,
\csc^{-1}
,
\sec^{-1}
, and
\cot^{-1}
, respectively.
Below, we will find the derivatives of arcsine, arccosine, and arctangent.
Derivative of arcsin
(f^{-1})'(y)=\frac{1}{f'(x)}
(\arcsin)'(y)=\frac{1}{\sin'(x)}
(\arcsin)'(y)=\frac{1}{\cos(x)}
Let's define the derivative in terms of
y
.
\sin(x)=y
\sin^2(x)+\cos^2(x)=1
\cos^2(x)=1-\sin^2(x)
\cos(x)=\sqrt{1-\sin^2(x)}
\cos(x)=\sqrt{1-y^2}
(\arcsin)'(y)=\frac{1}{\sqrt{1-y^2}}
Derivative of arccos
(f^{-1})'(y)=\frac{1}{f'(x)}
(\arccos)'(y)=\frac{1}{\cos'(x)}
(\arccos)'(y)=\frac{1}{-\sin(x)}
Let's define the derivative in terms of
y
.
\cos(x)=y
\sin^2(x)+\cos^2(x)=1
\sin^2(x)=1-\cos^2(x)
\sin(x)=\sqrt{1-\cos^2(x)}
\sin(x)=\sqrt{1-y^2}
(\arccos)'(y)=\frac{1}{-\sqrt{1-y^2}}
Derivative of arctan
(f^{-1})'(y)=\frac{1}{f'(x)}
(\arctan)'(y)=\frac{1}{\tan'(x)}
(\arctan)'(y)=\frac{1}{\sec^2(x)}
Let's define the derivative in terms of
y
.
\tan(x)= y
\sec^2(x)=\tan^2(x)+1
\sec^2(x)=y^2+1
(\arctan)'(y)=\frac{1}{y^2 + 1}
For a refresher on trig functions, check out
Appendix - Common Derivatives.