2.2 Trigonometric Functions and Their Inverses

The graphs of the six trigonometric functions are shown in Figure 2.2.1:

tikzpicture[scale=1.2,every node/.style=font=] scope[shift=(0,0),color=linecolor,line width=1.5pt,x=3cm/360] 0,5,...,360 sin( ) stroke [black!60,line width=1pt,-latex] (0,0) -- (400,0) node[right] x; [black!60,line width=1pt,-latex] (0,-1.2) -- (0,1.4) node[above] y; [black,left] at (0,0) 0; in 90,180,270,360 [black!60,line width=0.3pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in -1,1 [black!60,line width=0.3pt,shift=(0, )] (3pt,0pt) -- (-3pt,0pt); [black,left] at (0,1) 1; [black,left] at (0,-1) -1; [black,below] at (90,-0.1) 2; [black,below] at (180,-0.1) ; [black,below] at (270,-0.1) 3 2; [black,below] at (368,-0.1) 2 ; scope tikzpicture — Figure 2.2.1Graphs of the six trigonometric functions

tikzpicture[scale=1.2,every node/.style=font=] scope[shift=(0,0),color=linecolor,line width=1.5pt,x=3cm/360] 0,5,...,360 cos( ) stroke [black!60,line width=1pt,-latex] (0,0) -- (400,0) node[right] x; [black!60,line width=1pt,-latex] (0,-1.2) -- (0,1.4) node[above] y; [black,left] at (0,0) 0; in 90,180,270,360 [black!60,line width=0.3pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in -1,1 [black!60,line width=0.3pt,shift=(0, )] (3pt,0pt) -- (-3pt,0pt); [black,left] at (0,1) 1; [black,left] at (0,-1) -1; [black,below] at (90,-0.1) 2; [black,below] at (180,-0.1) ; [black,below] at (275,-0.1) 3 2; [black,below] at (362,-0.1) 2 ; scope tikzpicture — Figure 2.2.1Graphs of the six trigonometric functions

Recall that $\sin\,x$, $\cos\,x$, $\csc\,x$, and $\sec\,x$ have a period of $2\pi$ (i.e. repeat the same values every $2\pi$ radians), while $\tan\,x$ and $\cot\,x$ have a period of $\pi$.

The derivatives of the six trigonometric functions—given in Section 1.4—are:

Theorem 2.3

\begin{alignat*}{4} \ddx\,(\sin x) ~&=~ \cos x \qquad\qquad\qquad\qquad& \ddx\,(\csc x) ~&=~ -\csc x \; \cot x\\[6pt] \ddx\,(\cos x) ~&=~ -\sin x \qquad\qquad\qquad\qquad& \ddx\,(\sec x) ~&=~ \sec x \; \tan x\\[6pt] \ddx\,(\tan x) ~&=~ \sec^2 x \qquad\qquad\qquad\qquad& \ddx\,(\cot x) ~&=~ -\csc^2 x \end{alignat*}

The six trigonometric functions are not one-to-one over their entire domains, but recall from trigonometry that they are one-to-one when restricted to smaller domains, and hence have inverse functions, called the inverse trigonometric functions. For example, $y = \sin\,x$ is one-to-one over the interval $\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$, as shown in Figure 2.2.2 below:

tikzpicture[scale=1.2,every node/.style=font=] scope[black!60,dashed,line width=1.5pt,x=6cm/360] [black!60,solid,line width=1pt,-latex] (-200,0) -- (220,0) node[right] x; [black!60,solid,line width=1pt,-latex] (0,-1.2) -- (0,1.5) node[above] y; -180,-175,...,-90 sin( ) stroke 90,95,...,180 sin( ) stroke [black,below right] at (0,0) 0; in -180,-90,90,180 [black!60,line width=0.3pt,solid,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in -1,1 [black!60,line width=0.3pt,solid,shift=(0, )] (3pt,0pt) -- (-3pt,0pt) node[black,left] ; [black,below] at (90,-0.1) 2; [black,below] at (180,-0.1) ; [black,below] at (-90,-0.1) - 2; [black,below] at (-180,-0.1) - ; [black,above] at (180,1) y= \;x; scope scope[color=linecolor,line width=1.5pt,x=6cm/360] -90,-85,...,90 sin( ) stroke (-90,-1) circle (2pt); (90,1) circle (2pt); scope tikzpicture — Figure 2.2.2$y=\sin\,x$ is one-to-one with $x$ restricted to $\left[ -\frac{\pi}{2},\frac{\pi}{2} \right]$

Similarly, recall that $\cos\,x$ is one-to-one over $\ival{0}{\pi}$, $\tan\,x$ is one-to-one over $(-\pi/2,\pi/2)$, $\csc\,x$ is one-to-one over $(-\pi/2,0) \cup (0,\pi/2)$, $\sec\,x$ is one-to-one over $(0,\pi/2) \cup (\pi/2,\pi)$, and $\cot\,x$ is one-to-one over $(0,\pi)$. Hence, the inverse trigonometric functions $\sin^{-1} x$, $\cos^{-1} x$, $\tan^{-1} x$, $\csc^{-1} x$, $\sec^{-1} x$ and $\cot^{-1} x$ are defined,^[1] with the following domains and ranges:

1.5

function	$\sin^{-1} x$	$\cos^{-1} x$	$\tan^{-1} x$	$\csc^{-1} x$	$\sec^{-1} x$	$\cot^{-1} x$
domain	$\ival{-1}{1}$	$\ival{-1}{1}$	$(-\infty,\infty)$	$\abs{x} \ge 1$	$\abs{x} \ge 1$	$(-\infty,\infty)$
range	$\ival{-\tfrac{\pi}{2}}{\tfrac{\pi}{2}}$	$\ival{0}{\pi}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$	$\left(-\tfrac{\pi}{2},0\right) \cup \left(0,\tfrac{\pi}{2}\right)$	$\left(0,\tfrac{\pi}{2}\right) \cup \left(\tfrac{\pi}{2},\pi\right)$	$(0,\pi)$

The graphs of all six inverse trigonometric functions are shown in Figures 2.2.3 and 2.2.4 below:

tikzpicture[scale=1.2,every node/.style=font=] scope[x=6cm/360] [black!60,line width=1pt,-latex] (-70,0) -- (80,0) node[right] x; [black!60,line width=1pt,-latex] (0,-1.9) -- (0,2.1) node[above] y; [black,below right] at (0,0) 0; in -60,60 [black!60,line width=0.3pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in -1.5 [black!60,line width=0.3pt,shift=(0, )] (3pt,0pt) -- (-3pt,0pt) node[black,left] - 2; in 1.5 [black!60,line width=0.3pt,shift=(0, )] (3pt,0pt) -- (-3pt,0pt) node[black,left] 2; [black,below] at (60,-0.1) 1; [black,below] at (-60,-0.1) -1; scope scope[color=linecolor,line width=1.5pt,x=6cm/360,cm=0,1,1,0,(0,0)] -90,-85,...,90 sin( ) stroke [black] (-90,-1) circle (2pt); [black] (90,1) circle (2pt); scope tikzpicture — Figure 2.2.3Graphs of $\sin^{-1} x$, $\cos^{-1} x$, $\tan^{-1} x$

tikzpicture[scale=1.2,every node/.style=font=] scope[x=6cm/360] [black!60,line width=1pt,-latex] (-70,0) -- (90,0) node[right] x; [black!60,line width=1pt,-latex] (0,-0.2) -- (0,3.5) node[above] y; [black,below right] at (0,0) 0; in -60,60 [black!60,line width=0.3pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in 3 [black!60,line width=0.3pt,shift=(0, )] (3pt,0pt) -- (-3pt,0pt) node[black,left] ; [black,below] at (60,-0.1) 1; [black,below] at (-60,-0.1) -1; scope scope[color=linecolor,line width=1.5pt,x=6cm/360,cm=0,1,1,0,(0,0)] 0,5,...,180 cos( ) stroke [black] (0,1) circle (2pt); [black] (180,-1) circle (2pt); scope tikzpicture — Figure 2.2.3Graphs of $\sin^{-1} x$, $\cos^{-1} x$, $\tan^{-1} x$

tikzpicture[scale=0.8,every node/.style=font=] scope[line width=1pt,x=4cm/360] [black!60,solid,line width=1pt,-latex] (-230,0) -- (230,0) node[right] x; [black!60,solid,line width=1pt,-latex] (0,-2) -- (0,2.5) node[above] y; [black,above left] at (0,0) 0; in -90,90 [black!60,line width=0.3pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in 1.5 [black!60,line width=0.3pt,shift=(0, )] (-3pt,0pt) -- (3pt,0pt) node[black,left] 2; in -1.5 [black!60,line width=0.3pt,shift=(0, )] (-3pt,0pt) -- (3pt,0pt) node[black,left] - 2; [black,below] at (90,-0.1) 1; [black,below] at (-90,-0.1) -1; scope scope[color=linecolor,line width=1.5pt,x=4cm/360,cm=0,1,1,0,(0,0)] -135,-130,...,-70 1/sin(45+ ) 70,75,...,135 1/sin(-45+ ) stroke [black] (135,1) circle (2.5pt); [black] (-135,-1) circle (2.5pt); scope tikzpicture — Figure 2.2.4Graphs of $\csc^{-1} x$, $\sec^{-1} x$, $\cot^{-1} x$

tikzpicture[scale=0.8,every node/.style=font=] scope[line width=1.2pt,x=4cm/360] [black!60,-latex] (-230,0) -- (230,0) node[right] x; [black!60,-latex] (0,0) -- (0,3.8) node[above] y; [black,below] at (0,0) 0; in -90,90 [black!60,line width=0.4pt,shift=( ,0)] (0pt,3pt) -- (0pt,-3pt); in 3 [black!60,line width=0.4pt,shift=(0, )] (-3pt,0pt) -- (3pt,0pt) node[black,left] ; in 1.5 [black!60,line width=0.4pt,shift=(0, )] (-3pt,0pt) -- (3pt,0pt) node[black,left] 2; [black,below] at (90,-0.1) 1; [black,below] at (-90,-0.1) -1; [black!60,dashed,line width=0.4] (-230,1.5) -- (230,1.5); scope scope[color=linecolor,line width=1.5pt,x=4cm/360,cm=0,1,1,0,(0,0)] 0,5,...,65 1/cos( ) 205,210,...,270 1/cos(-90+ ) stroke [black] (270,-1) circle (2.5pt); [black] (0,1) circle (2.5pt); scope tikzpicture — Figure 2.2.4Graphs of $\csc^{-1} x$, $\sec^{-1} x$, $\cot^{-1} x$

The derivatives of the six inverse trigonometric functions are:

Theorem 2.4

\begin{alignat*}{4} \ddx\,(\sin^{-1} x) ~&=~ \frac{1}{\sqrt{1 - x^2}} \quad\text{(for $\abs{x} < 1$)} \qquad\quad\quad& \ddx\,(\csc^{-1} x) ~&=~ -\,\frac{1}{\abs{x}\sqrt{x^2 - 1}} \quad\text{(for $\abs{x} > 1$)}\\[6pt] \ddx\,(\cos^{-1} x) ~&=~ -\,\frac{1}{\sqrt{1 - x^2}} \quad\text{(for $\abs{x} < 1$)} \qquad\quad\quad& \ddx\,(\sec^{-1} x) ~&=~ \frac{1}{\abs{x}\sqrt{x^2 - 1}} \quad\text{(for $\abs{x} > 1$)}\\[6pt] \ddx\,(\tan^{-1} x) ~&=~ \frac{1}{1 + x^2} \qquad\quad\quad& \ddx\,(\cot^{-1} x) ~&=~ -\,\frac{1}{1 + x^2} \end{alignat*}

For the derivative of $\cos^{-1} x$, recall that $y = \cos^{-1} x$ is an angle between $0$ and $\pi$ radians, defined for $-1 \le x \le 1$. Since $\cos y = x$ by the definition of $y$, then $\dxdy = -\sin y$ and

\[\sin^2 y ~=~ 1 - \cos^2 y ~=~ 1 - x^2 \quad\Rightarrow\quad \sin y ~=~ \pm\,\sqrt{1 - x^2} ~=~ \sqrt{1 - x^2}\]

since $0 \le y \le \pi$ (which means $\sin y$ must be nonnegative). Thus:

\[\ddx\,(\cos^{-1} x) ~=~ \dydx ~=~ \frac{1}{\dxdy} ~=~ \frac{1}{-\sin y} ~=~ -\,\frac{1}{\sqrt{1 - x^2}} \quad\checkmark\]

For the derivative of $\sec^{-1} x$, since $y = \sec^{-1} x$ is defined for $\abs{x} \ge 1$, then $0 \le y < \pi/2$ for $x \ge 1$ and $\pi/2 < y \le \pi$ for $x \le -1$. Recall also that $\sec y$ and $\tan y$ are both positive when $0 < y < \pi/2$ and are both negative when $\pi/2 < y < \pi$. So in both cases the product $\sec y \; \tan y$ is nonnegative, i.e. $\sec y \; \tan y = \abs{\sec y \; \tan y}$. Thus, since $\sec y = x$ and

\[1 ~+~ \tan^2 y ~=~ \sec^2 y \quad\Rightarrow\quad \tan^2 y ~=~ \sec^2 y ~-~ 1 \quad\Rightarrow\quad \tan y ~=~ \pm\,\sqrt{\sec^2 y ~-~ 1} ~=~ \pm\,\sqrt{x^2 - 1}\]

then for $\abs{x} > 1$:

\[ \ddx\,(\sec^{-1} x) ~=~ \dydx ~=~ \frac{1}{\dxdy} ~=~ \frac{1}{\sec y \; \tan y} ~=~ \frac{1}{\abs{\sec y \; \tan y}} ~=~ \frac{1}{\Abs{x \sqrt{x^2 - 1}}} ~=~ \frac{1}{\abs{x}\sqrt{x^2 - 1}} \quad\checkmark \]

The proofs of the derivative formulas for the remaining inverse trigonometric functions are similar, and are left as exercises.

Example 2.3

Find the derivative of the function $y = 3\,\tan\,(\pi - 2x)$.

Solution: By the Chain Rule with $u = \pi - 2x$, the derivative of $y = 3\,\tan\,(\pi - 2x) = 3\,\tan u$ is:

\[\dydx ~=~ \dydu \;\cdot\; \dudx ~=~ \left(3\,\sec^2 u\right) \; (-2) ~=~ -6\,\sec^2\,(\pi - 2x)\]

Example 2.4

Find the derivative of the function $y = \sin^{-1}\,(x/4)$.

Solution: By the Chain Rule with $u = x/4$, the derivative of $y = \sin^{-1}\,(x/4) = \sin^{-1} u$ is:

\[ \dydx ~=~ \dydu \;\cdot\; \dudx ~=~ \frac{1}{\sqrt{1 - u^2}} \;\cdot\; \frac{1}{4} ~=~ \frac{1}{4\,\sqrt{1 - (x^2/16)}} ~=~ \frac{1}{\sqrt{16 - x^2}} \]

Exercises

For Exercises 1-16, find the derivative of the given function $y = f(x)$.

$y ~=~ \sec^2 3x$
$y ~=~ \csc (x^2 + 1)$
$y ~=~ \cot 3x$
$y ~=~ \cos\,(\tan x)$

4
$y ~=~ \tan^{-1} (x/3)$
$y ~=~ \sec^{-1} (x^2 + 1)$
$y ~=~ \cot^{-1} 3x$
$y ~=~ \cos^{-1}\,(\sin x)$

4
$y ~=~ \cot^{-1} (1/x)$
$y ~=~ \tan^{-1} \sqrt{x}$
$y ~=~ \left(\sin^{-1} 3x\right)^2$
$y ~=~ \tan^{-1} \frac{1}{x} + \tan^{-1} x$

4
$y ~=~ \tan^{-1} \frac{x-1}{x+1}$
$y ~=~ x\,\sin^{-1} (2x+1)$
$y ~=~ x\,\cot^{-1} x$
$y ~=~ \tan^{-1} \frac{1}{x} + \cot^{-1} x$
Find the derivative of $y = \sin^{-1} x ~+~ \cos^{-1} x \;$. Explain why no derivative formulas were needed.

For Exercises 18-21 prove the given derivative formula.

$\Ddx\,(\sin^{-1} x) ~=~ \dfrac{1}{\sqrt{1 - x^2}}$
$\Ddx\,(\tan^{-1} x) ~=~ \dfrac{1}{1 + x^2}$

2
$\Ddx\,(\cot^{-1} x) ~=~ -\,\dfrac{1}{1 + x^2}$
$\Ddx\,(\csc^{-1} x) ~=~ -\,\dfrac{1}{\abs{x}\sqrt{x^2 - 1}}$
The Chebyshev polynomials $T_n(x) = \cos\,(n\,\cos^{-1} x)$ are defined for all $\abs{x} \le 1$ and $n = 0, 1, 2, \ldots$.
1. Show that the Chebyshev polynomials $T_{n}(x)$ satisfy the differential equation
  
  \[ (1 - x^2)\;T_{n}''(x) ~-~ x\;T_{n}'(x) ~+~ n^2\,T_{n}(x) ~=~ 0 \]
2. Find polynomial expressions for $T_0(x)$, $T_1(x)$ and $T_2(x)$.
3. Show that $T_{n+1}(x) \;+\; T_{n-1}(x) \;=\; 2x\,T_{n}(x)$ for all $n \ge 1$. (Hint: Write $\theta = \cos^{-1} x$ so that $\cos \theta = x$)

The arc notation $\arcsin\,x$, $\arccos\,x$, $\arctan\,x$, $\arccsc\,x$, $\arcsec\,x$, $\arccot\,x$ is often used in place of $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$, $\csc^{-1}x$, $\sec^{-1}x$, $\cot^{-1}x$, respectively. ↩

function	\(\sin^{-1} x\)	\(\cos^{-1} x\)	\(\tan^{-1} x\)	\(\csc^{-1} x\)	\(\sec^{-1} x\)	\(\cot^{-1} x\)
domain	\(\ival{-1}{1}\)	\(\ival{-1}{1}\)	\((-\infty,\infty)\)	\(\abs{x} \ge 1\)	\(\abs{x} \ge 1\)	\((-\infty,\infty)\)
range	\(\ival{-\tfrac{\pi}{2}}{\tfrac{\pi}{2}}\)	\(\ival{0}{\pi}\)	\(\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\)	\(\left(-\tfrac{\pi}{2},0\right) \cup \left(0,\tfrac{\pi}{2}\right)\)	\(\left(0,\tfrac{\pi}{2}\right) \cup \left(\tfrac{\pi}{2},\pi\right)\)	\((0,\pi)\)