18b. Inverse Trigonometric Functions — Practice

Drill sheet for 18. Inverse Trig Functions. Book-style ideas rewritten as guided proofs and regenerable worksheets.

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1. Prove $D_x(\sec^{-1}x)=\dfrac{1}{x\sqrt{x^{2}-1}}$

Let $y=\sec^{-1}x$, so $\sec y=x$. Implicit differentiation:

$$\tan y\cdot\sec y\cdot y'=1.$$

$\tan^{2}y=\sec^{2}y-1=x^{2}-1$. On this chapter’s range ($[0,\pi/2)\cup[\pi,3\pi/2)$), $\tan y\ge 0$, so $\tan y=\sqrt{x^{2}-1}$. Hence

$$y'=\frac{1}{\tan y\cdot\sec y}=\frac{1}{x\sqrt{x^{2}-1}}.$$

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2–4. Chain rule with $\arcsin$, $\arccos$, $\arctan$

Book. $\arcsin(2x-3)$, $\arccos(x^{2})$, $\arctan(3x^{2})$.

$$\begin{aligned} D_x\arcsin(2x-3)&=\frac{2}{\sqrt{1-(2x-3)^{2}}},\\ D_x\arccos(x^{2})&=-\frac{2x}{\sqrt{1-x^{4}}},\\ D_x\arctan(3x^{2})&=\frac{6x}{1+9x^{4}}. \end{aligned}$$

Chain rule — arcsin / arccos / arctan

Outer inverse-trig derivative × inner

(1)Differentiate: (a) $y=\arcsin(2x-3)$ (b) $y=\arccos(x^{2})$ (c) $y=\arctan(3x^{2})$.

(a) $2/\sqrt{1-(2x-3)^{2}}$.

(b) $-2x/\sqrt{1-x^{4}}$.

(c) $6x/(1+9x^{4})$.

See steps.

(2)Differentiate $y=\arcsin(x^{3})$.

$y'=3x^{2}/\sqrt{1-x^{6}}$.

$y'=3x^{2}/\sqrt{1-x^{6}}$.

(3)Differentiate $y=\arccos(x^{2})$.

$y'=-2x/\sqrt{1-x^{4}}$.

$y'=-2x/\sqrt{1-x^{4}}$.

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5. $\cot^{-1}$ of a quotient

Book. $y=\cot^{-1}\big((1+x)/(1-x)\big)$ simplifies to $y'=-1/(1+x^{2})$.

cot⁻¹ / sec⁻¹ / csc⁻¹ compositions

Watch domain restrictions

(1)Differentiate $y=\cot^{-1}\big((1+x)/(1-x)\big)$.

Chain with (18.4); simplifies to $y'=-1/(1+x^{2})$.

$y'=-1/(1+x^{2})$.

(2)For $0<x<1$, simplify $y'$ if $y=x\csc^{-1}(1/x)+\sqrt{1-x^{2}}$.

$y'=\csc^{-1}(1/x)$.

$y'=\csc^{-1}(1/x)$.

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6–8. Combined / simplifying derivatives

Book 6. $y=x\sqrt{a^{2}-x^{2}}+a^{2}\arcsin(x/a)$ ⇒ $y'=2\sqrt{a^{2}-x^{2}}$.

Book 7. $y=x\csc^{-1}(1/x)+\sqrt{1-x^{2}}$ ($0<x<1$) ⇒ $y'=\csc^{-1}(1/x)$.

Book 8. $y=\dfrac{1}{ab}\arctan\big(\dfrac{b}{a}\tan x\big)$ ⇒ $y'=\dfrac{1}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}$.

Combined expressions that simplify

Expand carefully; cancel radicals

(1)If $y=\dfrac{1}{ab}\arctan\big(\dfrac{b}{a}\tan x\big)$, simplify $y'$.

$y'=1/(a^{2}\cos^{2}x+b^{2}\sin^{2}x)$.

$y'=1/(a^{2}\cos^{2}x+b^{2}\sin^{2}x)$.

(2)For $0<x<1$, if $y=x\csc^{-1}(1/x)+\sqrt{1-x^{2}}$, find $y'$.

$y'=\csc^{-1}(1/x)$.

$y'=\csc^{-1}(1/x)$.

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9. Implicit differentiation

Book. $y^{2}\sin x+y=\arctan x$:

$$y'=\frac{1-(1+x^{2})y^{2}\cos x}{(1+x^{2})(2y\sin x+1)}.$$

Implicit with inverse trig

Differentiate both sides; solve for y′

(1)If $y^{2}\sin x+y=\arctan x$, find $y'$.

$y'=\dfrac{1-(1+x^{2})y^{2}\cos x}{(1+x^{2})(2y\sin x+1)}$.

See steps.

(2)If $\arctan y=x^{2}$, find $y'$.

$y'=2x(1+y^{2})$.

$y'=2x(1+y^{2})$.

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10. Evaluate special values

Remember ranges: $\arcsin\in[-\pi/2,\pi/2]$, $\arccos\in[0,\pi]$, $\arctan\in(-\pi/2,\pi/2)$; this chapter’s $\sec^{-1}(2)=\pi/3$, $\sec^{-1}(-2)=4\pi/3$.

Evaluate inverse trig

Special angles + correct range

(1)Evaluate: (a) $\arcsin(-\sqrt{2}/2)$ (b) $\arccos(1/2)$ (c) $\arctan(-\sqrt{3})$ (d) $\sec^{-1}(2)$.

(a) $-\pi/4$; (b) $\pi/3$; (c) $-\pi/3$; (d) $\pi/3$.

$-\pi/4$; $\pi/3$; $-\pi/3$; $\pi/3$.

(2)Evaluate $\sec^{-1}(-2)$ (this chapter’s range).

$\sec(4\pi/3)=-2$ and $4\pi/3\in[\pi,3\pi/2)$.

$4\pi/3$.

(3)Evaluate $\arcsin(\sin\pi)$.

$\sin\pi=0$ ⇒ $\arcsin 0=0$ (not $\pi$).

$0$.

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11. Prove $\arcsin x+\arccos x=\pi/2$

Derivative is $0$ ⇒ constant on $(-1,1)$; at $x=0$ the value is $\pi/2$.

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12. $\arcsin(\sin x)$ vs the identity

(a) $\sin(\arcsin y)=y$ by definition (for $y\in[-1,1]$).

(b) $\arcsin(\sin\pi)=\arcsin 0=0$ — not $\pi$.

(c) $\arcsin(\sin x)=x$ iff $x\in[-\pi/2,\pi/2]$.

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13. Nested trig of inverse trig

Book. $\cos\big(2\arcsin(2/5)\big)=17/25$; $\sin\big(\arccos(-3/4)\big)=\sqrt{7}/4$ (QII ⇒ positive).

Compose trig with inverse trig

Use Pythagorean / double-angle identities

(1)Evaluate: (a) $\cos\big(2\arcsin(2/5)\big)$ (b) $\sin\big(\arccos(-3/4)\big)$.

(a) $1-2(2/5)^{2}=17/25$.

(b) $+\sqrt{7}/4$ (QII).

$17/25$; $\sqrt{7}/4$.

(2)Simplify $\cos\big(2\arcsin(2/5)\big)$.

$\cos 2\theta=1-2\sin^{2}\theta$ with $\sin\theta=2/5$.

$17/25$.

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14. Mural — maximize subtended angle

Mural height $12$ ft; bottom $6$ ft above eye. Distance $x$ from the wall:

$$\tan\theta=\frac{12x}{x^{2}+108},\qquad \theta=\arctan\!\left(\frac{12x}{x^{2}+108}\right).$$

$\theta'=0$ at $x=6\sqrt{3}\approx 10.4$ ft (relative max by the first-derivative test).

Fig. 18-7 — subtended angle θ

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Supplementary Problems

15. More special values

(a) $\arcsin(-\sqrt{3}/2)=-\pi/3$ · (b) $\arccos(\sqrt{3}/2)=\pi/6$ · (c) $\arccos(-\sqrt{3}/2)=5\pi/6$ · (d) $\arctan(-\sqrt{3}/3)=-\pi/6$ · (e) $\sec^{-1}\sqrt{2}=\pi/4$ · (f) $\sec^{-1}(-\sqrt{2})=5\pi/4$ (this chapter’s range).

Supplementary evaluations

Watch ranges for arccos and sec⁻¹

(1)Evaluate $\arccos(-\sqrt{3}/2)$ and $\sec^{-1}(-\sqrt{2})$.

$5\pi/6$; $5\pi/4$ (this chapter’s sec⁻¹ range).

$5\pi/6$; $5\pi/4$.

(2)Evaluate $\arccos(\cos(3\pi/2))$.

$\cos(3\pi/2)=0$ ⇒ $\arccos 0=\pi/2$.

$\pi/2$.

16. Prove $\arctan x+\mathrm{arccot}\,x=\pi/2$

Derivative $0$ ⇒ constant; at $x=0$ get $\pi/2$.

17–20. More chain-rule derivatives

Book. $\arcsin(3x)$, $\arccos(x/2)$, $\arctan(3/x)$, $\arcsin(x-1)$.

More chain-rule drills

Problems 17–20 style

(1)Differentiate: (a) $\arcsin(3x)$ (b) $\arccos(x/2)$ (c) $\arctan(3/x)$ (d) $\arcsin(x-1)$.

(a) $3/\sqrt{1-9x^{2}}$.

(b) $-1/\sqrt{4-x^{2}}$.

(c) $-3/(x^{2}+9)$.

(d) $1/\sqrt{2x-x^{2}}$.

See steps.

(2)Differentiate $y=\arcsin(3x)$.

$y'=3/\sqrt{1-9x^{2}}$.

$y'=3/\sqrt{1-9x^{2}}$.

(3)Differentiate $y=\arccos(x/2)$.

$y'=-1/\sqrt{4-x^{2}}$.

$y'=-1/\sqrt{4-x^{2}}$.

21–24. Products that simplify

Book answers.

• $y=x^{2}\arccos(2/x)$ ⇒ $y'=2x\big(\arccos(2/x)+1/\sqrt{x^{2}-4}\big)$
• $y=x/\sqrt{a^{2}-x^{2}}-\arcsin(x/a)$ ⇒ $y'=x^{2}/(a^{2}-x^{2})^{3/2}$
• $y=(x-a)\sqrt{2ax-x^{2}}+a^{2}\arcsin((x-a)/a)$ ⇒ $y'=2\sqrt{2ax-x^{2}}$
• $y=\sqrt{x^{2}-4}/x^{2}+\tfrac12\sec^{-1}(x/2)$ ⇒ $y'=8/(x^{3}\sqrt{x^{2}-4})$

Book OCR

Some printings write $\arcsin(x-a)$ in Prob. 22; the derivative answer matches $\arcsin(x/a)$.

Product / combined simplifications

Problems 21–24

(1)Differentiate $y=x^{2}\arccos(2/x)$ ($|x|>2$).

$y'=2x\big(\arccos(2/x)+1/\sqrt{x^{2}-4}\big)$.

$y'=2x\big(\arccos(2/x)+1/\sqrt{x^{2}-4}\big)$.

(2)If $y=(x-a)\sqrt{2ax-x^{2}}+a^{2}\arcsin((x-a)/a)$, show $y'=2\sqrt{2ax-x^{2}}$.

Differentiate; terms combine.

$y'=2\sqrt{2ax-x^{2}}$.

25. Prove (18.2), (18.4), (18.6)

Same implicit style as $(\sec^{-1})'$ and $(\arcsin)'$.

26–27. All six trig functions from one inverse

Book 26. $\theta=\arccos(2/7)$ ⇒ $\sin=3\sqrt{5}/7$, $\tan=3\sqrt{5}/2$, $\cos 2\theta=-41/49$, …

Book 27. $\theta=\arcsin(-1/5)$ ⇒ $\cos=+2\sqrt{6}/5$ (QIV), $\sin 2\theta=-4\sqrt{6}/25$, …

28. Double-angle for tangent

Prove $\tan 2\theta=\dfrac{2\tan\theta}{1-\tan^{2}\theta}$ from the addition formula (Ch. 16).

29. Nested compositions

(a) $\cos(\arcsin(3/11))=4\sqrt{7}/11$ · (b) $\tan(\sec^{-1}(7/5))=2\sqrt{6}/5$ · (c) expand $\sin(\arccos(2/5)+\sec^{-1}4)$ with angle-addition · (d) $\arccos(\cos(3\pi/2))=\pi/2$.

Book key

Some answer keys list $\tan(\sec^{-1}(7/5))$ as $2\sqrt{6}/7$ (that is $\sin$). The tangent is opp/adj $=2\sqrt{6}/5$.

Triangles from inverse values

Problems 26–29 / 35 style

(1)Let $\theta=\arccos(2/7)$. Find $\sin\theta$, $\tan\theta$, and $\cos 2\theta$.

$\sin=3\sqrt{5}/7$; $\tan=3\sqrt{5}/2$; $\cos 2\theta=-41/49$.

$3\sqrt{5}/7$; $3\sqrt{5}/2$; $-41/49$.

(2)Domain and range of $f(x)=\sin(\sec^{-1}x)$?

Domain $|x|\ge 1$; range $(-1,1)$.

Domain $|x|\ge 1$; range $(-1,1)$.

(3)Find $\sec(\arctan(5/7))$ and $\sec(\arctan(2x))$.

$\sec=\sqrt{74}/7$; formula $\sqrt{1+4x^{2}}$.

$\sqrt{74}/7$; $\sqrt{1+4x^{2}}$.

30–31. Domain / range and $\arctan(\tan x)$

$f(x)=\sin(\sec^{-1}x)$: domain $|x|\ge 1$, range $(-1,1)$.

$\arctan(\tan x)=x$ precisely on $(-\pi/2,\pi/2)$.

32. Light over a circular plot

Radius $30$ ft; $I=kx/(x^{2}+900)^{3/2}$. Maximize: $x=15\sqrt{2}$ ft.

Fig. 18-8 — maximize edge illumination

Illumination max

Problem 32

(1)Circular plot radius $30$ ft; $I=kx/(x^{2}+900)^{3/2}$. Maximize height $x$.

$I'=0$ ⇒ $900-2x^{2}=0$ ⇒ $x=15\sqrt{2}$ ft.

$15\sqrt{2}$ ft.

33 / 36. Relating inverse functions

$\arcsin x=\arctan\big(x/\sqrt{1-x^{2}}\big)$ for $|x|<1$ (limits at the endpoints).

For this chapter’s $\sec^{-1}$: $\sec^{-1}x=\arccos(1/x)$ when $x\ge 1$, and $\sec^{-1}x=2\pi-\arccos(1/x)$ when $x\le -1$.

34–35. Calculator / algebraic checks

$\arcsin(3/5)\approx 0.6435$ rad. $\sec(\arctan(5/7))=\sqrt{74}/7$; $\sec(\arctan(2x))=\sqrt{1+4x^{2}}$.

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Mixed review set

Full mixed review (1–36 style)

All inverse-trig skills

(1)For which $x$ is $\arctan(\tan x)=x$?

On $(-\pi/2,\pi/2)$.

$-\pi/2<x<\pi/2$.

(2)Circular plot radius $30$ ft; $I=kx/(x^{2}+900)^{3/2}$. Maximize height $x$.

$I'=0$ ⇒ $900-2x^{2}=0$ ⇒ $x=15\sqrt{2}$ ft.

$15\sqrt{2}$ ft.

(3)Differentiate $y=\arcsin(3x)$.

$y'=3/\sqrt{1-9x^{2}}$.

$y'=3/\sqrt{1-9x^{2}}$.

(4)Differentiate $y=\arctan(3/x)$.

$y'=-3/(x^{2}+9)$.

$y'=-3/(x^{2}+9)$.

(5)Differentiate $y=\cot^{-1}\big((1+x)/(1-x)\big)$.

After algebra: $y'=-1/(1+x^{2})$.

$y'=-1/(1+x^{2})$.

(6)If $y=\dfrac{1}{ab}\arctan\big(\dfrac{b}{a}\tan x\big)$, simplify $y'$.

$y'=1/(a^{2}\cos^{2}x+b^{2}\sin^{2}x)$.

$y'=1/(a^{2}\cos^{2}x+b^{2}\sin^{2}x)$.

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Self-check

Quiz

Inverse trig practice review

1 / 7

Dₓ(sec⁻¹ x) equals…

Pick an answer to see if you are right.

Further Reading

• 18. Inverse Trig Functions — theory and explorers
• 19. Rectilinear & Circular Motion — next: velocity and free fall
• 17b. Trig Diff — Practice — prior trig differentiation
• 17. Differentiation of Trig Functions — sin′ / tan′ / sec′
• 13. Law of the Mean — f′=0 ⇒ constant (used in Prob. 11)