Inverse trig graphs

A function has an inverse only if it is one-to-one. sin x on all of ℝ is not one-to-one — it repeats every 2π. So we restrict the domain. The restricted function has an inverse, and its graph is the reflection of the restricted original about the line y = x.

Restricted sin x → arcsin

We restrict sin x to [−π/2, π/2]. On this interval, sin x is strictly increasing from −1 to 1, so it's one-to-one.

Reflect this restricted piece about y = x to get arcsin:

• Domain of arcsin: [−1, 1]
• Range of arcsin: [−π/2, π/2]
• arcsin(0) = 0, arcsin(1) = π/2, arcsin(−1) = −π/2
• The graph is increasing, passes through the origin, bounded above by π/2 and below by −π/2
• It is concave down on [0, 1] and concave up on [−1, 0]

Restricted cos x → arccos

We restrict cos x to [0, π]. On this interval, cos x is strictly decreasing from 1 to −1.

Reflect about y = x:

• Domain of arccos: [−1, 1]
• Range of arccos: [0, π]
• arccos(1) = 0, arccos(0) = π/2, arccos(−1) = π
• The graph is strictly decreasing
• arcsin(x) + arccos(x) = π/2 for all x ∈ [−1, 1] — the two graphs are reflections of each other about the horizontal midline y = π/4

Restricted tan x → arctan

We restrict tan x to (−π/2, π/2). Tan is increasing on this open interval with vertical asymptotes at ±π/2.

Reflect about y = x:

• Domain of arctan: ℝ (all reals)
• Range of arctan: (−π/2, π/2)
• Horizontal asymptotes at y = π/2 (as x → +∞) and y = −π/2 (as x → −∞)
• arctan(0) = 0, arctan(1) = π/4, arctan(−1) = −π/4
• The graph has an S-shape with inflection at origin

The key identity chain

$\sin(\arcsin x) = x \quad \text{for } x \in [-1, 1]$ $\arcsin(\sin x) = x \quad \text{only for } x \in [-\pi/2, \pi/2]$

The second line is what students forget. arcsin(sin(5π/6)) ≠ 5π/6. Since 5π/6 is outside [−π/2, π/2], we need the equivalent angle inside that range: sin(5π/6) = sin(π/6) = 1/2, so arcsin(sin(5π/6)) = π/6.

Summary table

Function	Domain	Range	Increasing?
arcsin x	[−1, 1]	[−π/2, π/2]	Yes
arccos x	[−1, 1]	[0, π]	No
arctan x	ℝ	(−π/2, π/2)	Yes
arccot x	ℝ	(0, π)	No
arcsec x	(−∞,−1]∪[1,∞)	[0,π]{π/2}	Yes
arccosec x	(−∞,−1]∪[1,∞)	[−π/2,π/2]{0}	No

Drawing from scratch under exam pressure

You do not need to remember the shape in detail. Know:

1. The domain and range (from the restricted original)
2. Three anchor points (at the endpoints and at 0)
3. Whether increasing or decreasing
4. For arctan: horizontal asymptotes at ±π/2

Draw axes, mark anchor points, connect with a monotone curve that respects the asymptotes or bounds. That is enough for any JEE graph sketching question.