Class 12 Maths, Worked in Full Past papersGet updates
Chapter 2 · Inverse Trigonometric Functions

Exercise 2.1 · Q13 (MCQ): If sin⁡−1x=y\sin^{-1}x=y⁠, where can yy lie?

Exercise 2.1Principal values · Section 2.2
The question

If sin⁡−1x=y\sin^{-1}x = y⁠, then

(A) 0≤y≤π0 \le y \le \pi
(B) −π2≤y≤π2-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}
(C) 0<y<π0 < y < \pi
(D) −π2<y<π2-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}

Exam answer

Range of principal value branch of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

So sin⁡−1x=y\sin^{-1}x = y gives −π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2}⁠.

Answer: (B) −π2≤y≤π2-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The question asks for the range of sin⁡−1\sin^{-1}⁠, that is, the set of all values yy it can output. By definition, sin⁡−1\sin^{-1} is the inverse of sine restricted to [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, so its outputs are exactly the angles in that closed interval. Both ends are reached: sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2} and sin⁡−1(−1)=−π2\sin^{-1}(-1) = -\frac{\pi}{2}⁠.

Given, to find, and the plan
Given
y=sin⁡−1xy = \sin^{-1}x⁠, the principal value (no other branch is named).
To find
Which of (A)–(D) describes every possible value of yy⁠.
Plan
(1) Recall why sine is restricted to [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠. (2) Show that every yy in that interval occurs, ends included, and nothing outside it occurs. (3) Test each option with explicit values.

Basics you need

1An inverse needs a one-one, onto function

A function has an inverse only if it is one-one and onto. Sine on all of R\R is neither one-one nor onto R\R⁠, so its domain must be cut down first. Basics: what an inverse needs

2The range of the inverse is the restricted domain

If f:X→Yf : X \to Y is a bijection, then f−1:Y→Xf^{-1} : Y \to X⁠. So the outputs of sin⁡−1\sin^{-1} are exactly the angles of the interval to which sine was restricted (Section 2.2). Lesson: why the domain must be restricted

3sin on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

On this interval, sin increases steadily from sin⁡(−π2)=−1\sin\left(-\frac{\pi}{2}\right) = -1 to sin⁡π2=1\sin\frac{\pi}{2} = 1⁠. Basics: domain and range of sin and cos

4Closed and open intervals

a≤y≤ba \le y \le b includes both ends; a<y<ba < y < b excludes them. One value that a claim excludes but that actually occurs is enough to rule the claim out. Basics: angles and radians

Solution

  1. Where the range of sin⁡−1\sin^{-1} comes from

    Section 2.2 restricts sine to [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠. On this interval sine is one-one, and it takes every value in [−1,1][-1,1]⁠, so

    sin⁡:[−π2,π2]→[−1,1]\sin : \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \to [-1,1]

    is a bijection. Its inverse goes the other way:

    sin⁡−1:[−1,1]→[−π2,π2].\sin^{-1} : [-1,1] \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right].

    So whenever y=sin⁡−1xy = \sin^{-1}x⁠, yy is an angle in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, and every such angle occurs: given yy in the interval, take x=sin⁡y∈[−1,1]x = \sin y\in[-1,1]⁠; then sin⁡−1x=y\sin^{-1}x = y⁠.

    The book notes that other intervals, such as [π2,3π2]\left[\frac{\pi}{2},\frac{3\pi}{2}\right]⁠, would also make sine one-one; each gives a different branch. The branch with [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] is the principal value branch, and "sin⁡−1\sin^{-1}" with no branch named always means this one (Note, point 2).
    The range of an inverse function is the domain of the original. Lesson: branches

  2. The ends are included

    The ends ±π2\pm\frac{\pi}{2} are genuine values:

    sin⁡π2=1  ⇒  sin⁡−1(1)=π2,sin⁡(−π2)=−1  ⇒  sin⁡−1(−1)=−π2.\begin{aligned} \sin\tfrac{\pi}{2} &= 1 \;\Rightarrow\; \sin^{-1}(1) = \tfrac{\pi}{2},\\ \sin\left(-\tfrac{\pi}{2}\right) &= -1 \;\Rightarrow\; \sin^{-1}(-1) = -\tfrac{\pi}{2}. \end{aligned}

    So the range is the closed interval: −π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2}.

    O π/6 5π/6 −π/2 (x = −1) π/2 (x = 1) Principal range of sin⁻¹: −π/2 ≤ y ≤ π/2 (closed) Each x in [−1, 1] is the height of exactly one point of the green half. Example x = 1/2: blue π/6 is in range; rose 5π/6 is not. Both ends are filled: sin⁻¹(1) = π/2 and sin⁻¹(−1) = −π/2 are values.
    The values of sin⁡−1\sin^{-1} fill the closed right half of the unit circle (green), both ends included. Each height x∈[−1,1]x\in[-1,1] is met exactly once on this half; for x=12x = \frac12 that is π6\frac{\pi}{6} (blue), while 5π6\frac{5\pi}{6} (rose) has the same height but is not a value of sin⁡−1\sin^{-1}⁠.
    This is different from tan⁡−1\tan^{-1}⁠, whose range (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) is open because tan⁡(±π2)\tan\left(\pm\frac{\pi}{2}\right) does not exist. Sine is perfectly defined at ±π2\pm\frac{\pi}{2}⁠, so nothing forces those ends out.
    sin⁡π2=1\sin\frac{\pi}{2} = 1 and sin⁡(−π2)=−1\sin\left(-\frac{\pi}{2}\right) = -1⁠. Basics: sin and cos on the unit circle

  3. Examine every option

    OptionClaimVerdict
    (A)0≤y≤π0 \le y \le \piWrong. This is the range of cos⁡−1\cos^{-1}⁠. It misses negative values: sin⁡−1(−12)=−π6<0\sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6} < 0⁠. It also includes angles such as 2π3\frac{2\pi}{3} that sin⁡−1\sin^{-1} never outputs.
    (B)−π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2}Correct. Exactly the principal range found in Steps 1 and 2.
    (C)0<y<π0 < y < \piWrong. This is the range of cot⁡−1\cot^{-1}⁠. It excludes y=0y = 0⁠, yet sin⁡−1(0)=0\sin^{-1}(0) = 0⁠; and it excludes all negative values, such as −π6-\frac{\pi}{6}⁠.
    (D)−π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2}Wrong. This is the range of tan⁡−1\tan^{-1}⁠. It drops the ends, but sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2} and sin⁡−1(−1)=−π2\sin^{-1}(-1) = -\frac{\pi}{2} are values.
    Each wrong option is the range of a different inverse trig function, which is exactly why they are tempting. To rule an option out we only need one actual value of sin⁡−1\sin^{-1} that it excludes; the table gives one for each.
    Principal ranges (Section 2.2): sin⁡−1\sin^{-1}⁠: [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠; cos⁡−1\cos^{-1}⁠: [0,π][0,\pi]⁠; tan⁡−1\tan^{-1}⁠: (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠; cot⁡−1\cot^{-1}⁠: (0,π)(0,\pi)⁠. Lesson: principal values
Result

The correct answer is (B): if sin⁡−1x=y\sin^{-1}x = y⁠, then −π2≤y≤π2-\frac{\pi}{2} \le y \le \frac{\pi}{2}⁠. (Textbook answer: B.)

Check it with numbers

A calculator in radians gives sin⁡−1(1)≈1.5707963\sin^{-1}(1) \approx 1.5707963⁠, equal to π2\frac{\pi}{2}⁠, so the right end is attained ✓. sin⁡−1(−1)≈−1.5707963\sin^{-1}(-1) \approx -1.5707963⁠, the left end ✓. sin⁡−1(−0.5)≈−0.5235988\sin^{-1}(-0.5) \approx -0.5235988⁠, a negative value, so (A) and (C) fail ✓. sin⁡−1(0.999)≈1.5260712\sin^{-1}(0.999) \approx 1.5260712⁠, just inside the range, never beyond π2\frac{\pi}{2} ✓.

Common mistakes
  1. Choosing (D) by mixing up sin⁡−1\sin^{-1} with tan⁡−1\tan^{-1}⁠. Only tan⁡−1\tan^{-1} has the open interval.
  2. Choosing (A) by mixing up sin⁡−1\sin^{-1} with cos⁡−1\cos^{-1}⁠.
  3. Thinking yy can be any angle with sin⁡y=x\sin y = x⁠. That describes the equation sin⁡y=x\sin y = x⁠, not the function sin⁡−1\sin^{-1}⁠, which outputs exactly one angle.
What to take away
  1. The range of sin⁡−1\sin^{-1} is the closed interval [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, the restricted domain of sine.
  2. A point is left out of a principal range only where the original function is undefined: the ends ±π2\pm\frac{\pi}{2} for tan⁡−1\tan^{-1}⁠, the ends 00 and π\pi for cot⁡−1\cot^{-1}⁠, and an inside point for sec⁡−1\sec^{-1} (π2\frac{\pi}{2}⁠) and cosec⁡−1\operatorname{cosec}^{-1} (00⁠), whose ends stay in.
  3. To eliminate an MCQ option about a range, one actual output outside the claimed set is enough.

Every fact used, and where it comes from

FactUsed inSource
Sine restricted to [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] is one-one and onto [−1,1][-1,1]Step 1Section 2.2 (Lesson: why the domain must be restricted)
Inverse of a bijection f:X→Yf : X \to Y is f−1:Y→Xf^{-1} : Y \to XStep 1Section 2.2, recalling Chapter 1
Principal branch sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1,1] \to \left[-\frac{\pi}{2},\frac{\pi}{2}\right]Steps 1–3Section 2.2 and its table of principal value branches
No branch named means the principal branchStep 1Note, point 2, Section 2.2
Ranges of cos⁡−1\cos^{-1}⁠, cot⁡−1\cot^{-1}⁠, tan⁡−1\tan^{-1}Step 3Table of principal value branches, Section 2.2
sin⁡−1(−12)=−π6\sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}Step 3Exercise 2.1 Q1
All numerical valuesCheckRecomputed in the site's verification script
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →