Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

WHAT: the definitions

HOW we pick the ranges (derivation from first principles)

Rule: to invert a function you need it to be one-to-one (bijective) on the chosen piece. So pick the simplest continuous interval that (a) hits every output value once and (b) includes $x=0$ nicely.

Sine increases on $[-\pi/2, \pi/2]$ and there covers all outputs $[-1,1]$ exactly once. ✅ Pick it.
Cosine on $[-\pi/2,\pi/2]$ is not one-to-one ( $\cos(-a)=\cos a$ ). The next simplest monotone interval covering $[-1,1]$ once is $[0,\pi]$ (decreasing). ✅ Pick it.
Tangent increases on $(-\pi/2,\pi/2)$ and there sweeps from $-\infty$ to $+\infty$ . Endpoints are asymptotes → open interval.

Figure — Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

Formula Handy identities (derived, not memorised)

arcsin + arccos: Let $\theta=\arcsin x$ , so $\sin\theta=x$ with $\theta\in[-\pi/2,\pi/2]$ . Then $\cos(\pi/2-\theta)=\sin\theta=x$ , and $\pi/2-\theta\in[0,\pi]$ (the valid arccos range). Hence $\arcsin x + \arccos x = \frac{\pi}{2}, \quad x\in[-1,1].$ Oddness: $\sin$ is odd so its inverse is odd: $\arcsin(-x)=-\arcsin x$ , and $\arctan(-x)=-\arctan x$ . But $\arccos(-x)=\pi-\arccos x$ (from the identity above).

Worked examples

Forecast-then-Verify

Recall Predict before revealing

Q: What is $\arcsin(\sin\,(3\pi/4))$ ? First guess " $3\pi/4$ "... Verify: $3\pi/4\notin[-\pi/2,\pi/2]$ , so the output can't be $3\pi/4$ . Compute $\sin(3\pi/4)=\sqrt2/2$ , then $\arcsin(\sqrt2/2)=\pi/4$ . Answer: $\pi/4$ . The inverse "snaps back" into the range.

Common mistakes (Steel-manned)

Flashcards

Domain of arcsin

[-1,1]

Range of arcsin

[-\pi/2,\ \pi/2]

Range of arccos

[0,\ \pi]

Domain of arctan

all real numbers

(-\infty,\infty)

Range of arctan

(-\pi/2,\ \pi/2)

(open — asymptotes)

Why must we restrict sine's domain to define arcsin?

Sine isn't one-to-one; restricting to

[-\pi/2,\pi/2]

makes each ratio map to a unique angle.

\arcsin x + \arccos x = ?

\pi/2

for

x\in[-1,1]

\arccos(-x)

in terms of

\arccos x

\pi-\arccos x

\cos(\arcsin x)=?

\sqrt{1-x^2}

\arcsin(\sin(3\pi/4))=?

\pi/4

(snap into range)

Why is an inverse graph a reflection across

y=x

Because

(a,b)\in f \iff (b,a)\in f^{-1}

\arcsin(\sin x)=x

always true?

No — only for

x\in[-\pi/2,\pi/2]

Recall Feynman: explain to a 12-year-old

A machine turns an angle into a "slant number." Sometimes you know the slant number and want to know which angle made it. That's the inverse machine. Trouble: many angles make the same slant number (like a clock at 30° and 150° both leaning the same for sine). So we tell the inverse machine: "only ever answer with angles from this small safe zone." That zone is the range. Now every answer is unique and never confusing.

Connections

Unit circle and reference angles
Trigonometric identities
One-to-one functions and invertibility
Derivatives of inverse trig functions
Solving trigonometric equations
Reflection of graphs across y=x

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, normal $\sin$ , $\cos$ , $\tan$ functions angle ko lete hain aur ek ratio dete hain. Inverse trig ulta kaam karta hai — ratio do, angle wapas milega. Lekin ek problem hai: $\sin 30°$ aur $\sin 150°$ dono $0.5$ hote hain, matlab ek hi ratio ke liye bohot saare angles. To agar hum aise hi inverse banaye to answer confusing ho jayega. Isliye hum function ka domain chota kar dete hain ek aise piece pe jaha function strictly badh ya ghat raha ho — is chote piece ko hi inverse ka range kehte hain.

Yaad rakhne wali baat: arcsin ka range $[-\pi/2, \pi/2]$ , arccos ka $[0,\pi]$ , aur arctan ka $(-\pi/2, \pi/2)$ (open, kyunki $\pi/2$ pe asymptote hai — kabhi touch nahi karta). Domain me arcsin aur arccos sirf $[-1,1]$ lete hain (ratio 1 se zyada ho hi nahi sakta), jabki arctan har real number leta hai.

Graph ka trick simple hai: inverse ka graph original ka reflection hota hai $y=x$ line ke across. Kyunki agar $(a,b)$ original pe hai to $(b,a)$ inverse pe hoga. Aur ek super useful identity: $\arcsin x + \arccos x = \pi/2$ — isko rata mat, derive karo $\cos(\pi/2 - \theta) = \sin\theta$ se.

Sabse common galti: students sochte hain $\arcsin(\sin x) = x$ har baar. Nahi! Ye tabhi sahi hai jab $x$ pehle se range $[-\pi/2,\pi/2]$ me ho. Warna pehle ratio nikalo, phir range me "snap" karo. Exam me yahi trap aata hai, to hamesha range check karo.

Inverse trig functions — arcsin, arccos, arctan — domain, range, graphs

WHAT: the definitions

HOW we pick the ranges (derivation from first principles)

Worked examples

Forecast-then-Verify

Common mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Advanced Trigonometry

Connections