2.12: Inverse Trigonometric Functions
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One very useful application of implicit differentiation is to find the derivatives of inverse functions. We have already used this approach to find the derivative of the inverse of the exponential function — the logarithm.
We are now going to consider the problem of finding the derivatives of the inverses of trigonometric functions. Now is a very good time to go back and reread Section 0.6 on inverse functions — especially Definition 0.6.4. Most importantly, given a function \(f(x)\text{,}\) its inverse function \(f^{-1}(x)\) only exists, with domain \(D\text{,}\) when \(f(x)\) passes the “horizontal line test”, which says that for each \(Y\) in \(D\) the horizontal line \(y=Y\) intersects the graph \(y=f(x)\) exactly once. (That is, \(f(x)\) is a one-to-one function.)
Let us start by playing with the sine function and determine how to restrict the domain of \(\sin x\) so that its inverse function exists.
Let \(y=f(x)=\sin(x)\text{.}\) We would like to find the inverse function which takes \(y\) and returns to us a unique \(x\)-value so that \(\sin(x)=y\text{.}\)
- For each real number \(Y\text{,}\) the number of \(x\)-values that obey \(\sin(x)=Y\text{,}\) is exactly the number of times the horizontal straight line \(y=Y\) intersects the graph of \(\sin(x)\text{.}\)
- When \(-1\le Y\le 1\text{,}\) the horizontal line intersects the graph infinitely many times. This is illustrated in the figure above by the line \(y=0.3\text{.}\)
- On the other hand, when \(Y \lt -1\) or \(Y \gt 1\text{,}\) the line \(y=Y\) never intersects the graph of \(\sin(x)\text{.}\) This is illustrated in the figure above by the line \(y=-1.2\text{.}\)
This is exactly the horizontal line test and it shows that the sine function is not one-to-one.
Now consider the function
\begin{align*} y &= \sin(x) & \text{with domain } -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}
This function has the same formula but the domain has been restricted so that, as we'll now show, the horizontal line test is satisfied.
As we saw above when \(|Y| \gt 1\) no \(x\) obeys \(\sin(x)=Y\) and, for each \(-1\le Y\le 1\text{,}\) the line \(y=Y\) (illustrated in the figure above with \(y=0.3\)) crosses the curve \(y=\sin(x)\) infinitely many times, so that there are infinitely many \(x\)'s that obey \(f(x)=\sin x=Y\text{.}\) However exactly one of those crossings (the dot in the figure) has \(-\frac{\pi}{2}\le x \le\frac{\pi}{2}\text{.}\)
That is, for each \(-1\le Y \le 1\text{,}\) there is exactly one \(x\text{,}\) call it \(X\text{,}\) that obeys both
\begin{align*} \sin X &= Y &\text{and} && -\frac{\pi}{2}\le X \le \frac{\pi}{2} \end{align*}
That unique value, \(X\text{,}\) is typically denoted \(\arcsin(Y)\text{.}\) That is
\begin{align*} \sin( \arcsin(Y) ) &= Y & \text{and} && -\frac{\pi}{2}\le \arcsin(Y) \le\frac{\pi}{2} \end{align*}
Renaming \(Y\rightarrow x\text{,}\) the inverse function \(\arcsin(x)\) is defined for all \(-1 \le x \le 1\) and is determined by the equation
\[ \sin\big(\arcsin(x)\big)=x\qquad\text{and}\qquad -\frac{\pi}{2}\le \arcsin(x)\le\frac{\pi}{2}. \nonumber \]
Note that many texts will use \(\sin^{-1}(x)\) to denote arcsine, however we will use \(\arcsin(x)\) since we feel that it is clearer 1; the reader should recognise both.
Since
\[ \sin\frac{\pi}{2}=1\qquad\sin\frac{\pi}{6}=\frac{1}{2} \nonumber \]
and \(-\frac{\pi}{2}\le \frac{\pi}{6},\frac{\pi}{2}\le \frac{\pi}{2}\text{,}\) we have
\[ \arcsin 1= \frac{\pi}{2}\qquad \arcsin \frac{1}{2}= \frac{\pi}{6} \nonumber \]
Even though
\[ \sin(2\pi)=0 \nonumber \]
it is not true that \(\arcsin 0 =2\pi\text{,}\) and it is not true that \(\arcsin\big(\sin(2\pi)\big) =2\pi\text{,}\) because \(2\pi\) is not between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\text{.}\) More generally
\begin{align*} \arcsin\big(\sin(x)\big) &=\text{ the unique angle } \theta \text{ between } -\frac{\pi}{2} \text{ and } \frac{\pi}{2} \text{ obeying } \sin \theta =\sin x\\ &= x\quad\text{if and only if $-\frac{\pi}{2}\le x\le \frac{\pi}{2}$} \end{align*}
So, for example, \(\arcsin\big(\sin\big(\frac{11\pi}{16}\big)\big)\) cannot be \(\frac{11\pi}{16}\) because \(\frac{11\pi}{16}\) is bigger than \(\frac{\pi}{2}\text{.}\) So how do we find the correct answer? Start by sketching the graph of \(\sin(x)\text{.}\)
It looks like the graph of \(\sin x\) is symmetric about \(x=\frac{\pi}{2}\text{.}\) The mathematical way to say that “the graph of \(\sin x\) is symmetric about \(x=\frac{\pi}{2}\)” is “\(\sin(\frac{\pi}{2}-\theta)= \sin(\frac{\pi}{2}+\theta)\)” for all \(\theta\text{.}\) That is indeed true 2.
Now \(\frac{11\pi}{16}=\frac{\pi}{2} +\frac{3\pi}{16}\) so
\[ \sin\Big(\frac{11\pi}{16}\Big) =\sin\Big(\frac{\pi}{2}+\frac{3\pi}{16}\Big) =\sin\Big(\frac{\pi}{2}-\frac{3\pi}{16}\Big) =\sin\Big(\frac{5\pi}{16}\Big) \nonumber \]
and, since \(\frac{5\pi}{16}\) is indeed between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\text{,}\)
\[ \arcsin\Big(\sin\Big(\frac{11\pi}{16}\Big)\Big) =\frac{5\pi}{16}\qquad\Big(\text{and not $\frac{11\pi}{16}$}\Big). \nonumber \]
Derivatives of Inverse Trig Functions
Now that we have explored the arcsine function we are ready to find its derivative. Lets call
\begin{align*} \arcsin(x) &= \theta(x), \end{align*}
so that the derivative we are seeking is \(\dfrac{d\theta}{dx}\text{.}\) The above equation is (after taking sine of both sides) equivalent to
\begin{align*} \sin(\theta) &= x \end{align*}
Now differentiate this using implicit differentiation (we just have to remember that \(\theta\) varies with \(x\) and use the chain rule carefully):
\begin{align*} \cos(\theta) \cdot \dfrac{d\theta}{dx} &= 1\\ \dfrac{d\theta}{dx} &= \frac{1}{\cos(\theta)} & \text{substitute $\theta = \arcsin x$}\\ \dfrac{d}{dx} \arcsin x &= \frac{1}{\cos(\arcsin x)} \end{align*}
This doesn't look too bad, but it's not really very satisfying because the right hand side is expressed in terms of \(\arcsin(x)\) and we do not have an explicit formula for \(\arcsin(x)\text{.}\)
However even without an explicit formula for \(\arcsin(x)\text{,}\) it is a simple matter to get an explicit formula for \(\cos\big(\arcsin(x)\big)\text{,}\) which is all we need. Just draw a right–angled triangle with one angle being \(\arcsin(x)\text{.}\) This is done in the figure below 3.
Since \(\sin(\theta)=x\) (see 2.12.2), we have made the side opposite the angle \(\theta\) of length \(x\) and the hypotenuse of length \(1\text{.}\) Then, by Pythagoras, the side adjacent to \(\theta\) has length \(\sqrt{1-x^2}\) and so
\begin{gather*} \cos\big(\arcsin(x)\big)=\cos(\theta)=\sqrt{1-x^2} \end{gather*}
which in turn gives us the answer we need:
\begin{gather*} \dfrac{d}{dx} \arcsin(x) =\frac{1}{\sqrt{1-x^2}} \end{gather*}
The definitions for \(\arccos\text{,}\) \(\arctan\) and \(\textrm{arccot}\) are developed in the same way. Here are the graphs that are used.
The definitions for the remaining two inverse trigonometric functions may also be developed in the same way 4 In fact, there are two different widely used definitions of \(\textrm{arcsec} x\text{.}\) Under our definition, below, \(\theta=\textrm{arcsec} x\) takes values in \(0\le\theta\le\pi\text{.}\) Some people, perfectly legitimately, define \(\theta=\textrm{arcsec} x\) to take values in the union of \(0\le \theta\lt\frac{\pi}{2}\) and \(\pi\le\theta\lt\frac{3\pi}{2}\text{.}\) Our definition is sometimes called the “trigonometry friendly” definition. The definition itself has the advantage of simplicity. The other definition is sometimes called the “calculus friendly” definition. It eliminates some absolute values and hence simplifies some computations. Similarly there are two different widely used definitions of \(\textrm{arccsc} x\text{.}\) 5 One could also define \(\textrm{arccot}(x)=\arctan(1/x)\) with \(\textrm{arccot}(0)=\frac{\pi}{2}\text{.}\) We have chosen not to do so, because the definition we have chosen is both continuous and standard.. But it's a little easier to use
\begin{gather*} \csc x=\frac{1}{\sin x} \qquad \sec x=\frac{1}{\cos x} \end{gather*}
\(\arcsin x\) is defined for \(|x|\le 1\text{.}\) It is the unique number obeying
\begin{align*} \sin\big(\arcsin(x)\big)&=x &&\text{and}& -\frac{\pi}{2}\le &\arcsin(x)\le\frac{\pi}{2}\\ \end{align*}
\(\arccos x\) is defined for \(|x|\le 1\text{.}\) It is the unique number obeying
\begin{align*} \cos\big(\arccos(x)\big)&=x &&\text{and}& 0\le &\arccos(x)\le\pi\\ \end{align*}\(\arctan x\) is defined for all \(x\in\mathbb{R}\text{.}\) It is the unique number obeying
\begin{align*} \tan\big(\arctan(x)\big)&=x &&\text{and}& -\frac{\pi}{2} \lt &\arctan(x) \lt \frac{\pi}{2}\\ \end{align*}\(\textrm{arccsc} x=\arcsin\frac{1}{x}\) is defined for \(|x|\ge 1\text{.}\) It is the unique number obeying
\begin{align*} \csc\big(\textrm{arccsc}(x)\big)&=x &&\text{and}& -\frac{\pi}{2}\le &\textrm{arccsc}(x)\le\frac{\pi}{2}\\ \end{align*}\(\ \ \ \ \ \ \ \ \)Because \(\csc(0)\) is undefined, \(\textrm{arccsc}(x)\) never takes the value \(0\text{.}\)
\begin{align*} \end{align*}\(\textrm{arcsec} x=\arccos\frac{1}{x}\) is defined for \(|x|\ge 1\text{.}\) It is the unique number obeying
\begin{align*} \sec\big(\textrm{arcsec}(x)\big)&=x &&\text{and}& 0\le &\textrm{arcsec}(x)\le\pi\\ \end{align*}\(\ \ \ \ \ \ \ \ \)Because \(\sec(\pi/2)\) is undefined, \(\textrm{arcsec}(x)\) never takes the value \(\pi/2\text{.}\)
\begin{align*} \end{align*}\(\textrm{arccot} x\) is defined for all \(x\in\mathbb{R}\text{.}\) It is the unique number obeying
\begin{align*} \cot\big(\textrm{arccot}(x)\big)&=x &&\text{and}& 0 \lt &\textrm{arccot}(x) \lt \pi \end{align*}To find the derivative of \(\arccos\) we can follow the same steps:
- Write \(\arccos(x) =\theta(x)\) so that \(\cos\theta = x\) and the desired derivative is \(\dfrac{d\theta}{dx}\text{.}\)
- Differentiate implicitly, remembering that \(\theta\) is a function of \(x\text{:}\)
\begin{align*} -\sin\theta \dfrac{d\theta}{dx} &= 1\\ \dfrac{d\theta}{dx} &= -\frac{1}{\sin\theta}\\ \dfrac{d}{dx}\arccos x &= -\frac{1}{\sin(\arccos x)}. \end{align*}
- To simplify this expression, again draw the relevant triangle
from which we see
\begin{align*} \sin(\arccos x) = \sin\theta &= \sqrt{1-x^2}. \end{align*}
- Thus
\begin{align*} \dfrac{d}{dx}\arccos x &= -\frac{1}{\sqrt{1-x^2}}. \end{align*}
Very similar steps give the derivative of \(\arctan x\text{:}\)
- Start with \(\theta = \arctan x\text{,}\) so \(\tan \theta = x\text{.}\)
- Differentiate implicitly:
\begin{align*} \sec^2 \theta \dfrac{d\theta}{dx} &= 1\\ \dfrac{d\theta}{dx} &= \frac{1}{\sec^2 \theta} = \cos^2 \theta\\ \dfrac{d}{dx}\arctan x &= \cos^2(\arctan x). \end{align*}
- To simplify this expression, we draw the relevant triangle
from which we see
\begin{gather*} \cos^2(\arctan x) = \cos^2\theta = \frac{1}{1+x^2} \end{gather*}
- Thus
\begin{align*} \dfrac{d}{dx}\arctan x &= \frac{1}{1+x^2}. \end{align*}
An almost identical computation gives the derivative of \(\textrm{arccot} x\text{:}\)
- Start with \(\theta = \textrm{arccot} x\text{,}\) so \(\cot \theta = x\text{.}\)
- Differentiate implicitly:
\begin{align*} -\csc^2 \theta \dfrac{d\theta}{dx} &= 1\\ \dfrac{d}{dx}\textrm{arccot} x = \dfrac{d\theta}{dx} &= -\frac{1}{\csc^2 \theta} = -\sin^2 \theta = -\frac{1}{1+x^2} \end{align*}
from the triangle
To find the derivative of \(\textrm{arccsc}\) we can use its definition and the chain rule.
\begin{align*} \theta &= \textrm{arccsc} x & \text{take cosecant of both sides}\\ \csc \theta &= x & \text{but $\csc \theta = \frac{1}{\sin\theta}$, so flip both sides}\\ \sin \theta &= \frac{1}{x} & \text{now take arcsine of both sides}\\ \theta &= \arcsin\left(\frac{1}{x}\right) \end{align*}
Now just differentiate, carefully using the chain rule :
\begin{align*} \dfrac{d\theta}{dx} &= \dfrac{d}{dx} \arcsin\left(\frac{1}{x}\right)\\ &= \frac{1}{\sqrt{1-x^{-2}}} \cdot \frac{-1}{x^2}\\ \end{align*}
To simplify further we will factor \(x^{-2}\) out of the square root. We need to be a little careful doing that. Take another look at examples 1.5.6 and 1.5.7 and the discussion between them before proceeding.
\begin{align*} &= \frac{1}{\sqrt{x^{-2}(x^2-1)}} \cdot \frac{-1}{x^2}\\ &= \frac{1}{|x^{-1}|\cdot \sqrt{x^2-1}} \cdot \frac{-1}{x^2} & \text{note that $x^2 \cdot |x^{-1}| = |x|$.}\\ &= - \frac{1}{|x|\sqrt{x^2-1}} \end{align*}In the same way we can find the derivative of the remaining inverse trig function. We just use its definition, a derivative we already know and the chain rule.
\begin{gather*} \dfrac{d}{dx} \textrm{arcsec}(x) = \dfrac{d}{dx} \arccos\Big(\frac{1}{x}\Big) =-\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot\Big(-\frac{1}{x^2}\Big) =\frac{1}{|x|\sqrt{x^2-1}} \end{gather*}
By way of summary, we have
The derivatives of the inverse trigonometric functions are
\begin{align*} \dfrac{d}{dx} \arcsin(x) &= \frac{1}{\sqrt{1-x^2}} & \dfrac{d}{dx} \textrm{arccsc}(x) &= -\frac{1}{|x|\sqrt{x^2-1}}\\ \dfrac{d}{dx} \arccos(x) &= -\frac{1}{\sqrt{1-x^2}} & \dfrac{d}{dx} \textrm{arcsec}(x) &= \frac{1}{|x|\sqrt{x^2-1}}\\ \dfrac{d}{dx} \arctan(x) &= \frac{1}{1+x^2} & \dfrac{d}{dx} \textrm{arccot}(x) &= -\frac{1}{1+x^2} \end{align*}
Exercises
Stage 1
Give the domains of each of the following functions.
\begin{align*} \mbox{(a) } f(x)&=\arcsin(\cos x) & \mbox{(b) } g(x)&=\textrm{arccsc}(\cos x)\\ \mbox{(c) } h(x)&=\sin(\arccos x) \end{align*}
A particle starts moving at time \(t=10\text{,}\) and it bobs up and down, so that its height at time \(t \geq 10\) is given by \(\cos t\text{.}\) True or false: the particle has height 1 at time \(t=\arccos(1)\text{.}\)
The curve \(y=f(x)\) is shown below, for some function \(f\text{.}\) Restrict \(f\) to the largest possible interval containing \(0\) over which it is one--to--one, and sketch the curve \(y=f^{-1}(x)\text{.}\)
Let \(a\) be some constant. Where does the curve \(y=ax+\cos x\) have a horizontal tangent line?
Define a function \(f(x)=\arcsin x + \textrm{arccsc} x\text{.}\) What is the domain of \(f(x)\text{?}\) Where is \(f(x)\) differentiable?
Stage 2
Differentiate \(f(x)=\arcsin\left(\dfrac{x}{3}\right)\text{.}\) What is the domain of \(f(x)\text{?}\)
Differentiate \(f(t)=\dfrac{\arccos t}{t^2-1}\text{.}\) What is the domain of \(f(t)\text{?}\)
Differentiate \(f(x)=\textrm{arcsec}(-x^2-2)\text{.}\) What is the domain of \(f(x)\text{?}\)
Differentiate \(f(x)=\dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right)\text{,}\) where \(a\) is a nonzero constant. What is the domain of \(f(x)\text{?}\)
Differentiate \(f(x)=x\arcsin x + \sqrt{1-x^2}\text{.}\) What is the domain of \(f(x)\text{?}\)
For which values of \(x\) is the tangent line to \(y=\arctan (x^2)\) horizontal?
Evaluate \(\displaystyle\dfrac{d}{dx}\{\arcsin x + \arccos x\}\text{.}\)
Find the derivative of \(y=\arcsin \!\big(\frac{1}{x}\big)\text{.}\)
Find the derivative of \(y=\arctan \big(\frac{1}{x}\big)\text{.}\)
Calculate and simplify the derivative of \((1+x^2)\arctan x\text{.}\)
Show that \(\displaystyle\dfrac{d}{dx}\left\{\sin\left(\arctan(x) \right)\right\} = (x^2+1)^{-3/2}\text{.}\)
Show that \(\displaystyle\dfrac{d}{dx}\left\{\cot\left(\arcsin(x) \right)\right\} = \dfrac{-1}{x^2\sqrt{1-x^2}}\text{.}\)
Determine all points on the curve \(y=\arcsin x\) where the tangent line is parallel to the line \(y=2x+9\text{.}\)
For which values of \(x\) does the function \(f(x)=\arctan(\csc x)\) have a horizontal tangent line?
Stage 3
Let \(f(x) = x + \cos x\text{,}\) and let \(g(y) = f^{-1}(y)\) be the inverse function. Determine \(g'(y)\text{.}\)
\(f(x) = 2x-\sin(x)\) is one--to--one. Find \(\big(f^{-1}\big)'(\pi-1)\text{.}\)
\(f(x) = e^x+x\) is one--to--one. Find \(\big(f^{-1}\big)'(e+1)\text{.}\)
Differentiate \(f(x)=[\sin x +2]^{\textrm{arcsec} x}\text{.}\) What is the domain of this function?
Suppose you can't remember whether the derivative of arcsine is \(\dfrac{1}{\sqrt{1-x^2}}\) or \(\dfrac{1}{\sqrt{x^2-1}}\text{.}\) Describe how the domain of arcsine suggests that one of these is wrong.
Evaluate \(\displaystyle \lim_{x\to 1}\left( (x-1)^{-1}\left(\arctan x - \frac{\pi}{4}\right)\right).\)
Suppose \(f(2x+1)=\dfrac{5x-9}{3x+7}\text{.}\) Evaluate \(f^{-1}(7)\text{.}\)
Suppose \(f^{-1}(4x-1)=\dfrac{2x+3}{x+1}\text{.}\) Evaluate \(f(0)\text{.}\)
Suppose a curve is defined implicitly by
\[ \arcsin(x+2y)=x^2+y^2 \nonumber \]
Solve for \(y'\) in terms of \(x\) and \(y\text{.}\)