MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: inverse trigonometric functions (MATH 203)

Q: Why is arcsin(sin x) not always equal to x?

Because arcsin can only return an angle between minus pi over 2 and pi over 2. When x is already in that interval, arcsin(sin x) gives x back. When it is not, arcsin returns the angle of that interval with the same sine: for x = 5 pi over 6, the sine is one half and arcsin gives pi over 6, not 5 pi over 6.

Q: Why does the derivative of arcsec x have an absolute value?

Because the graph of arcsec rises on both of its branches, so its slope is always positive, including for x less than minus 1. The absolute value appears in the computation when the square root of x squared is simplified: it equals the absolute value of x, not x. Without it, the formula would give negative slopes on the left branch.

Q: How do I find cos(arctan x) without a calculator?

Call the angle theta, so tan theta = x. Draw a right triangle with opposite side x and adjacent side 1; the hypotenuse is the square root of 1 plus x squared. So cos theta is 1 over that square root. The sign is positive because arctan returns angles between minus pi over 2 and pi over 2, where the cosine is never negative.

Q: How do you prove an identity with inverse trig functions using derivatives?

Show that the difference of the two sides has a zero derivative at every point of an interval; then it is constant on that interval. Read the constant at one convenient point of the same interval. If the domain has several intervals, do it once per interval: the constants can differ, as for arcsec and arctan of the square root of x squared minus 1.

Q: Do I need radians for inverse trig derivatives on a MATH 203 test?

Yes. The formulas such as the derivative of arctan x being 1 over 1 plus x squared are true only when angles are measured in radians. Set the calculator to radian mode before the test and check it: the inverse sine of 1 must display 1.5708, not 90.

This sheet is not a summary of sections 1.6 and 3.9 of Thomas: you already have the textbook. It answers one question only, what makes students lose marks on the inverse trigonometric functions in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The formulas of the chapter fit on one line each, and that is the danger: the marks are lost in the range that decides a sign, and in the algebra around the formula, a fraction inside a square root or the square root of a square. A scientific calculator is allowed, in radian mode; exact values are still expected where they exist.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

An inverse trigonometric function returns ONE angle, the one in its restricted range, and that range decides every sign of the chapter: $\sin^{-1}(\sin x)$ , the sides of the reference triangle, the $|x|$ of $\frac{d}{dx}\sec^{-1}x$ , the constant of an identity on each interval. The marks are lost in the algebra around it: $1 \pm u^2$ brought to one fraction, and $\sqrt{x^2} = |x|$ .

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Inverse sine function
Inverse cosine function
Inverse tangent function
Inverse secant function
Restricted domain and range
Reference right triangle
Derivatives of inverse trigonometric functions
Chain rule
Implicit differentiation
Functions with zero derivative
Corner point and differentiability

The essentials

One angle, the one in the range

• $y = \sin^{-1}x$ ( $= \arcsin x$ ) is the angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin y = x$ , for $-1 \le x \le 1$ . $y = \cos^{-1}x$ is the angle of $[0, \pi]$ with $\cos y = x$ . $y = \tan^{-1}x$ is the angle of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan y = x$ , for every $x$ .
• Thomas' $\sec^{-1}x = \cos^{-1}\frac{1}{x}$ , for $|x| \ge 1$ : an angle of $[0, \pi]$ , never $\frac{\pi}{2}$ .
• A negative input sends $\sin^{-1}$ and $\tan^{-1}$ below $0$ , and sends $\cos^{-1}$ and $\sec^{-1}$ into the second quadrant, $\left(\frac{\pi}{2}, \pi\right]$ .
• $\sin(\sin^{-1}x) = x$ on $[-1, 1]$ ; $\sin^{-1}(\sin x) = x$ ONLY on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ .
• Trig of an inverse: draw the triangle for the sizes, use the range for the signs.

\arcsin

lives between

-\frac{\pi}{2}

and

\frac{\pi}{2}

\arccos

between

0

and

\pi

: at

x = -1

one returns

-\frac{\pi}{2}

, the other

\pi

The $-1$ of $\sin^{-1}$ names an inverse function. The reciprocal $\frac{1}{\sin x}$ is $\csc x$ .

The derivatives, and the algebra that comes with them

• $\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1 - u^2}}$ , $\frac{d}{dx}\cos^{-1}u = -\frac{u'}{\sqrt{1 - u^2}}$ for $|u| < 1$ ; $\frac{d}{dx}\tan^{-1}u = \frac{u'}{1 + u^2}$ .
• $\frac{d}{dx}\sec^{-1}u = \frac{u'}{|u|\sqrt{u^2 - 1}}$ for $|u| > 1$ ; $\cot^{-1}$ and $\csc^{-1}$ have the opposite derivatives of $\tan^{-1}$ and $\sec^{-1}$ .
• Square the WHOLE inner function: $(3x)^2 = 9x^2$ , $(x^2)^2 = x^4$ , $(\sqrt x)^2 = x$ .
• Bring $1 \pm u^2$ to one fraction before the root or the division: $1 + \frac{9}{x^2} = \frac{x^2 + 9}{x^2}$ .
• $\sqrt{x^2} = |x|$ , $\sqrt{\sin^2 x} = |\sin x|$ , $\sqrt{(1 - x^2)^2} = |1 - x^2|$ : the root of a square is an absolute value.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Six inverse functions: range and derivative

Read a line as: the function of the first column returns an angle of the second, and has the derivative of the third. The red lines are formulas that students write and that are false on part of the domain.

Function	Range	Derivative
$\sin^{-1}x$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$	$\frac{1}{\sqrt{1 - x^2}}$
Example: At $x = \frac{1}{2}$ : $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ , slope $\frac{2}{\sqrt 3} \approx 1.1547$ .
$\cos^{-1}x$	$[0, \pi]$	$-\frac{1}{\sqrt{1 - x^2}}$
Example: $\cos^{-1}\left(-\frac{\sqrt 3}{2}\right) = \frac{5\pi}{6}$ , never $-\frac{\pi}{6}$ .
$\tan^{-1}x$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$	$\frac{1}{1 + x^2}$
Example: $\tan^{-1}(-1) = -\frac{\pi}{4}$ ; slope $\frac{1}{2}$ at $x = -1$ .
$\sec^{-1}x$	$[0, \pi]$ , not $\frac{\pi}{2}$	$\frac{1}{\|x\|\sqrt{x^2 - 1}}$
Example: $\sec^{-1}(-2) = \frac{2\pi}{3}$ , slope $\frac{1}{2\sqrt 3} \approx 0.2887 > 0$ .
$\sec^{-1}x$	copied from the right branch	$\frac{1}{x\sqrt{x^2 - 1}}$ false for x < -1
Example: At $x = -2$ it gives $-0.2887$ , a negative slope on a rising branch. What to do: Keep $\|x\|$ : it comes from $\sqrt{x^2} = \|x\|$ in $\cos^{-1}\frac{1}{x}$ .
$\sin^{-1}(3x)$	square on x only	$\frac{3}{\sqrt{1 - 3x^2}}$ no such rule
Example: At $x = 0.3$ it gives $3.511$ ; the true slope is $\frac{3}{\sqrt{0.19}} \approx 6.8825$ . What to do: $u = 3x$ , so $u^2 = 9x^2$ : $\frac{3}{\sqrt{1 - 9x^2}}$ , for $\|x\| < \frac{1}{3}$ .

Test of the sign: $\sin^{-1}$ , $\tan^{-1}$ , $\sec^{-1}$ rise on their whole domain, so their derivatives are positive; $\cos^{-1}$ , $\cot^{-1}$ , $\csc^{-1}$ fall.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Giving back the angle you started from

the whole question

What not to write

“ $\sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \frac{5\pi}{6}$ .”

What to write

“ $\sin\frac{5\pi}{6} = \frac{1}{2}$ , and the angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with sine $\frac{1}{2}$ is $\frac{\pi}{6}$ : $\sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}$ .”

Why: $\sin^{-1}$ can only return an angle of its range, and $\frac{5\pi}{6} > \frac{\pi}{2}$ . Compute from the inside out, then ask which angle OF THE RANGE.

2. The wrong sign in the reference triangle

2 marks

What not to write

“ $\tan^{-1}(-2)$ is negative, so $\cos(\tan^{-1}(-2)) = -\frac{1}{\sqrt 5}$ .”

What to write

“ $\theta = \tan^{-1}(-2)$ is in $\left(-\frac{\pi}{2}, 0\right)$ , where the cosine is positive: $\cos\theta = \frac{1}{\sqrt 5}$ and $\sin\theta = -\frac{2}{\sqrt 5}$ .”

Why: The triangle gives the sizes, $1$ , $2$ , $\sqrt 5$ ; the range gives the signs. A fourth-quadrant angle has a positive cosine and a negative sine.

3. Answering in degrees

the whole tangent line

What not to write

“ $\tan^{-1}1 = 45$ , so the tangent to $y = \tan^{-1}x$ at $x = 1$ is $y = 45 + \frac{1}{2}(x - 1)$ .”

What to write

“ $\tan^{-1}1 = \frac{\pi}{4} \approx 0.7854$ , so the tangent is $y = \frac{\pi}{4} + \frac{1}{2}(x - 1)$ .”

Why: The derivative formulas of section 3.9 are true in radians only. Check the calculator mode before the test: $\sin^{-1}(1)$ must display $1.5708$ .

4. Writing the square root of a square without absolute value

2 to 3 marks, and a formula false on a whole branch

What not to write

“ $\sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x}$ , so $\frac{d}{dx}\sec^{-1}x = \frac{1}{x\sqrt{x^2 - 1}}$ .”

What to write

“ $\sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{\sqrt{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$ , so $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}$ .”

x = -2

the left branch of

\sec^{-1}

rises, with slope

\frac{1}{2\sqrt 3} \approx 0.29

: the formula without

|x|

would give

-0.29

Why: $\sqrt{4} = 2$ , not $-2$ : for $x < 0$ the root of $x^2$ is $-x$ . The graph settles it, since both branches of $\sec^{-1}$ rise.

5. Squaring half of the inner function

2 marks

What not to write

“The domain of $\sin^{-1}(3x)$ is $[-1, 1]$ and its derivative is $\frac{3}{\sqrt{1 - 3x^2}}$ .”

What to write

“ $-1 \le 3x \le 1$ gives the domain $\left[-\frac{1}{3}, \frac{1}{3}\right]$ , and $u^2 = 9x^2$ gives $\frac{3}{\sqrt{1 - 9x^2}}$ .”

\sin^{-1}(3x)

covers the same heights as

\sin^{-1}x

on an interval three times shorter,

\left[-\frac{1}{3}, \frac{1}{3}\right]

Why: Both conditions are on the INNER function $u = 3x$ , never on $x$ alone. A derivative that exists at $x = 0.5$ , where the function does not, is the signal.

6. Losing the chain factor of a fraction

2 marks, and the wrong sign

What not to write

“ $\frac{d}{dx}\tan^{-1}\frac{1}{x} = \frac{1}{1 + \frac{1}{x^2}} = \frac{x^2}{x^2 + 1}$ .”

What to write

“ $\frac{d}{dx}\tan^{-1}\frac{1}{x} = \frac{-1/x^2}{1 + 1/x^2} = -\frac{1}{x^2 + 1}$ , after multiplying top and bottom by $x^2$ .”

Why: The formula is $\frac{u'}{1 + u^2}$ , and here $u' = -\frac{1}{x^2}$ . At $x = 1$ the student says $+\frac{1}{2}$ for a function that decreases.

7. One constant for two intervals

half of the question

What not to write

“ $g(x) = \tan^{-1}\sqrt{x^2 - 1}$ and $\sec^{-1}x$ have derivatives of the same size, so $\sec^{-1}x = \tan^{-1}\sqrt{x^2 - 1}$ for all $|x| \ge 1$ .”

What to write

“For $x > 1$ , $(g - \sec^{-1})' = 0$ and the constant, read at $x = 2$ , is $0$ . For $x < -1$ , $(g + \sec^{-1})' = 0$ and the constant, read at $x = -2$ , is $\pi$ .”

Why: A zero derivative gives a constant on ONE interval. The domain $|x| \ge 1$ has two pieces, each with its own constant, read at a point of that piece: $\sec^{-1}(-2) = \frac{2\pi}{3}$ , not $\frac{\pi}{3}$ .

8. Moving a minus sign through arccos

the whole value

What not to write

“ $\cos^{-1}(-0.2) = -\cos^{-1}(0.2) \approx -1.3694$ .”

What to write

“ $\cos^{-1}(-0.2) = \pi - \cos^{-1}(0.2) \approx 1.7722$ .”

Why: $\cos^{-1}$ never returns a negative angle. The minus sign goes out of $\sin^{-1}$ and $\tan^{-1}$ , which are odd, but not out of $\cos^{-1}$ : $\cos^{-1}(-x) = \pi - \cos^{-1}x$ .

Which method to choose

Which move, by the form of the question

Read the expression from the outside in before computing anything

If an inverse of a number: $\cos^{-1}\left(-\frac{\sqrt 2}{2}\right)$ → find the angle of the RANGE with that value

Example: $\cos^{-1}\left(-\frac{\sqrt 2}{2}\right) = \frac{3\pi}{4}$ , in $[0, \pi]$
If an inverse of a trig function: $\tan^{-1}(\tan a)$ → compute the inside, then the angle of the range; the answer is a only if a is already in the range

Example: $\tan^{-1}\left(\tan\frac{5\pi}{4}\right) = \frac{\pi}{4}$
If a trig function of an inverse: $\sin(\cos^{-1}t)$ → draw the triangle for the sizes, then the sign from the range

Example: $\sin\left(\cos^{-1}\left(-\frac{4}{5}\right)\right) = \frac{3}{5}$ , positive on $[0, \pi]$
If the derivative of $\sin^{-1}u$ , $\tan^{-1}u$ or $\sec^{-1}u$ → name $u$ , write $u'$ , apply the formula, then bring $1 \pm u^2$ to one fraction

Example: $\frac{d}{dx}\sin^{-1}\frac{x}{2} = \frac{1/2}{\sqrt{(4 - x^2)/4}} = \frac{1}{\sqrt{4 - x^2}}$
If a root of a square appears: $\sqrt{u^2}$ → write $|u|$ , then split the domain by the sign of $u$

Example: $\sqrt{\sin^2 x} = |\sin x|$ : $\sin x$ on $(0, \pi)$ , $-\sin x$ on $(-\pi, 0)$
If prove that two expressions are equal, or that one is constant → show a zero derivative on EACH interval of the domain, and read each constant at a point of its interval

Example: $\sin^{-1}x + \cos^{-1}x$ : derivative $0$ on $(-1, 1)$ , value $\frac{\pi}{2}$ at $x = 0$

No branch applies to an angle outside every range, as in $\cos^{-1}x = \frac{4\pi}{3}$ : that equation has no solution, and saying so is the answer.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving the derivative of arcsec by implicit differentiation

When to use it: Any question that says 'prove' or 'derive' the formula for $\frac{d}{dx}\sec^{-1}x$

1 Rewrite $y = \sec^{-1}x$ as $\sec y = x$ AND state the range: $0 \le y \le \pi$ , $y \ne \frac{\pi}{2}$ , for $|x| > 1$ .
2 Differentiate both sides with respect to $x$ , naming the chain rule: $\sec y\tan y \cdot y' = 1$ .
3 Write $\tan y = \pm\sqrt{x^2 - 1}$ from $\tan^2 y = \sec^2 y - 1$ , and choose the sign ON EACH BRANCH by the range.
4 Show that $\sec y\tan y = |x|\sqrt{x^2 - 1}$ in both cases, and conclude with the domain $|x| > 1$ .

Concluding sentence

“Let $y = \sec^{-1}x$ , so $\sec y = x$ with $y \in [0, \pi]$ , $y \ne \frac{\pi}{2}$ . Differentiating, $\sec y\tan y \cdot y' = 1$ . For $x > 1$ , $y \in \left(0, \frac{\pi}{2}\right)$ and $\tan y = \sqrt{x^2 - 1}$ ; for $x < -1$ , $y \in \left(\frac{\pi}{2}, \pi\right)$ and $\tan y = -\sqrt{x^2 - 1}$ . In both cases $\sec y\tan y = |x|\sqrt{x^2 - 1}$ , so $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}$ for $|x| > 1$ .”

The trap: Keeping the plus sign on both branches, which gives $\frac{1}{x\sqrt{x^2 - 1}}$ , negative for $x < -1$ .

Marking: Typically 1 mark for the equation with its range, 1 for the implicit differentiation, 2 for the sign on each branch, 1 for the final formula with its domain.

Proving an identity with a zero derivative

When to use it: 'Show that $F(x) = G(x)$ for...' or 'show that $F$ is constant' on a set made of one or several intervals

1 Write the domain as a union of INTERVALS, and name them.
2 Compute $(F - G)'$ and simplify it to $0$ on each interval, showing the algebra (one fraction, the absolute values).
3 Invoke the fact: a function with zero derivative at every point of an interval is constant on that interval.
4 Read the constant at one convenient point OF EACH interval, then check the endpoints directly.

Concluding sentence

“On the interval $(1, \infty)$ , $(F - G)'(x) = 0$ at every point, so $F - G$ is constant there. At $x = 2$ , a point of this interval, $F(2) - G(2) = 0$ . Hence $F(x) = G(x)$ for every $x > 1$ .”

The trap: Reading one constant at x = 0 and extending it to every interval of the domain.

Marking: Typically 2 marks for the zero derivative with its algebra, 1 for naming the interval, 1 for the constant read at a point of it, 1 for the endpoints.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The answer is in the range

An inverse sine or tangent between minus pi over 2 and pi over 2, an inverse cosine or secant between 0 and pi. An answer outside is wrong, whatever its sine or cosine.

$\cos^{-1}(-0.2) = -1.3694$ fails at once: a negative angle is not a value of $\cos^{-1}$ .
The sign of the derivative matches the graph

arcsin, arctan and arcsec rise, so their derivatives, and those of increasing inner functions composed with them, are positive; arccos falls.

$\frac{d}{dx}\sec^{-1}x$ at $x = -2$ must be positive: $\frac{1}{2\sqrt 3}$ passes, $-\frac{1}{2\sqrt 3}$ fails.
A symmetric difference quotient on the calculator

Compute (f(a + 0.001) minus f(a minus 0.001)) divided by 0.002 in radian mode. It must agree with your derivative to about four digits.

For $\sin^{-1}(3x)$ at $0.3$ : the quotient gives $6.88$ , which confirms $\frac{3}{\sqrt{0.19}} \approx 6.8825$ and refutes $3.511$ .
The identity at a second point

After reading a constant at one point of an interval, test the identity at another point of the SAME interval with the calculator.

$\sec^{-1}(-3) = \pi - \tan^{-1}\sqrt 8 \approx 1.9106$ , and $\cos^{-1}\left(-\frac{1}{3}\right) \approx 1.9106$ too.

The typical problem, taken apart

A derivative with a root of a square, and the corner it reveals

Let $y = \cos^{-1}(1 - 2x^2)$ for $-1 \le x \le 1$ . Find $y'$ for $x \ne 0$ , $|x| < 1$ , simplified, and describe the graph at $x = 0$ .

Every step must be justified as on a MATH 203 final.

The graph of

y = \cos^{-1}(1 - 2x^2)

is a V with its tip at the origin: the derivative must explain that tip.

Step 1

Inner function $u = 1 - 2x^2$ , $u' = -4x$ . Chain rule: $y' = -\frac{u'}{\sqrt{1 - u^2}} = \frac{4x}{\sqrt{1 - (1 - 2x^2)^2}}$ .

Why

Naming $u$ and $u'$ first earns the method mark and keeps the two minus signs from cancelling by accident.

Step 2

$1 - (1 - 2x^2)^2 = 1 - (1 - 4x^2 + 4x^4) = 4x^2 - 4x^4 = 4x^2(1 - x^2)$ .

Why

Expand, then FACTOR before taking the root: a root of a product splits, a root of a sum does not.

Step 3

$\sqrt{4x^2(1 - x^2)} = 2|x|\sqrt{1 - x^2}$ , since $\sqrt{x^2} = |x|$ .

Why

This is the line where the marks go: writing $2x$ instead of $2|x|$ makes the formula false for every negative $x$ .

Step 4

$y' = \frac{4x}{2|x|\sqrt{1 - x^2}}$ : for $0 < x < 1$ , $y' = \frac{2}{\sqrt{1 - x^2}}$ ; for $-1 < x < 0$ , $y' = -\frac{2}{\sqrt{1 - x^2}}$ .

Why

$\frac{x}{|x|}$ is $1$ or $-1$ according to the sign of $x$ : split the domain there and give one formula per piece.

Step 5

As $x \to 0^+$ , $y' \to 2$ ; as $x \to 0^-$ , $y' \to -2$ . The graph has a corner at the origin, and $y$ is not differentiable at $0$ .

Why

Two different one-sided slopes are the definition of a corner, and they match the V of the figure.

Step 6

Check: on $(0, 1)$ , $y - 2\sin^{-1}x$ has derivative $0$ , and at $x = \frac{1}{2}$ , $\cos^{-1}\frac{1}{2} = \frac{\pi}{3} = 2 \cdot \frac{\pi}{6}$ : so $y = 2\sin^{-1}x$ there.

Why

A second route to the same function confirms the derivative, and reads the constant at a point of the right interval.

The conclusion, written out

“ $y' = \frac{2}{\sqrt{1 - x^2}}$ for $0 < x < 1$ and $y' = -\frac{2}{\sqrt{1 - x^2}}$ for $-1 < x < 0$ . The one-sided slopes at $0$ are $2$ and $-2$ , so the graph has a corner at the origin and $y$ is not differentiable there.”

The classic mistake on this problem: Writing $\sqrt{4x^2(1 - x^2)} = 2x\sqrt{1 - x^2}$ : the answer $\frac{2}{\sqrt{1 - x^2}}$ everywhere claims a slope of about $2.31$ at $x = -\frac{1}{2}$ , where the graph goes DOWN.

Learn by heart

• $\sin^{-1}$ and $\tan^{-1}$ return angles of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (endpoints included for $\sin^{-1}$ ); $\cos^{-1}$ and $\sec^{-1}$ return angles of $[0, \pi]$ .
• $\sin^{-1}(\sin x) = x$ ONLY on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\cos^{-1}(-x) = \pi - \cos^{-1}x$ .
• Triangle for the sizes, range for the signs.
• $(\sin^{-1}u)' = \frac{u'}{\sqrt{1 - u^2}}$ , $(\tan^{-1}u)' = \frac{u'}{1 + u^2}$ , $(\sec^{-1}u)' = \frac{u'}{|u|\sqrt{u^2 - 1}}$ ; $\cos^{-1}$ , $\cot^{-1}$ , $\csc^{-1}$ : the same with a minus sign.
• $\sqrt{x^2} = |x|$ : the root of a square is an absolute value.
• Zero derivative on an interval: constant on THAT interval, read at a point of it.
• Radian mode, always.

Frequently asked questions

Why is arcsin(sin x) not always equal to x?

Because arcsin can only return an angle between minus pi over 2 and pi over 2. When x is already in that interval, arcsin(sin x) gives x back. When it is not, arcsin returns the angle of that interval with the same sine: for x = 5 pi over 6, the sine is one half and arcsin gives pi over 6, not 5 pi over 6.

Why does the derivative of arcsec x have an absolute value?

Because the graph of arcsec rises on both of its branches, so its slope is always positive, including for x less than minus 1. The absolute value appears in the computation when the square root of x squared is simplified: it equals the absolute value of x, not x. Without it, the formula would give negative slopes on the left branch.

How do I find cos(arctan x) without a calculator?

Call the angle theta, so tan theta = x. Draw a right triangle with opposite side x and adjacent side 1; the hypotenuse is the square root of 1 plus x squared. So cos theta is 1 over that square root. The sign is positive because arctan returns angles between minus pi over 2 and pi over 2, where the cosine is never negative.

How do you prove an identity with inverse trig functions using derivatives?

Show that the difference of the two sides has a zero derivative at every point of an interval; then it is constant on that interval. Read the constant at one convenient point of the same interval. If the domain has several intervals, do it once per interval: the constants can differ, as for arcsec and arctan of the square root of x squared minus 1.

Do I need radians for inverse trig derivatives on a MATH 203 test?

Yes. The formulas such as the derivative of arctan x being 1 over 1 plus x squared are true only when angles are measured in radians. Set the calculator to radian mode before the test and check it: the inverse sine of 1 must display 1.5708, not 90.

Practise it

Corrected exercises: Inverse trigonometric functions, MATH 203 at Concordia

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: inverse trigonometric functions (MATH 203)

The essentials

One angle, the one in the range

The derivatives, and the algebra that comes with them

The rules in table form

Six inverse functions: range and derivative

The mistakes that cost marks

1. Giving back the angle you started from

2. The wrong sign in the reference triangle

3. Answering in degrees

4. Writing the square root of a square without absolute value

5. Squaring half of the inner function

6. Losing the chain factor of a fraction

7. One constant for two intervals

8. Moving a minus sign through arccos

Which method to choose

Which move, by the form of the question

How the answer is expected to be written

Proving the derivative of arcsec by implicit differentiation

Proving an identity with a zero derivative

Check before you hand in

The typical problem, taken apart

A derivative with a root of a square, and the corner it reveals

Learn by heart

Frequently asked questions

Why is arcsin(sin x) not always equal to x?

Why does the derivative of arcsec x have an absolute value?

How do I find cos(arctan x) without a calculator?

How do you prove an identity with inverse trig functions using derivatives?

Do I need radians for inverse trig derivatives on a MATH 203 test?

Corrected exercises: Inverse trigonometric functions, MATH 203 at Concordia

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