Class 12 Maths, Worked in Full Past papersGet updates
CBSE Board Examination 2024 · Section E · 4 marks

Q38: Case study: the inverse of the sine function

Board 2024Ch 2 · Inverse Trigonometric FunctionsCh 1 · Relations and Functions
The question

If a function f:X→Yf : X \to Y defined as f(x)=yf(x) = y is one-one and onto, then we can define a unique function g:Y→Xg : Y \to X such that g(y)=xg(y) = x⁠, where x∈Xx \in X and y=f(x)y = f(x)⁠, y∈Yy \in Y⁠. Function g is called the inverse of function f.

The domain of sine function is R and function sine : R→RR \to R is neither one-one nor onto. The following graph shows the sine function.

5π 2 − π 2 − 3π 2 π 2 5π 2 −2π −π π 2π O 1 −1 X′ X Y Y′
y=sin⁡xy = \sin x

Let sine function be defined from set A to [−1,1][-1, 1] such that inverse of sine function exists, i.e., sin⁡−1x\sin^{-1}x is defined from [−1,1][-1, 1] to A.

On the basis of the above information, answer the following questions :

(i) If A is the interval other than principal value branch, give an example of one such interval. (1 mark)

(ii) If sin⁡−1(x)\sin^{-1}(x) is defined from [−1,1][-1, 1] to its principal value branch, find the value of sin⁡−1(−12)−sin⁡−1(1)\sin^{-1}\left(-\dfrac12\right) - \sin^{-1}(1)⁠. (1 mark)

(iii) (a) Draw the graph of sin⁡−1x\sin^{-1}x from [−1,1][-1, 1] to its principal value branch. (2 marks)

OR

(iii) (b) Find the domain and range of f(x)=2sin⁡−1(1−x)f(x) = 2\sin^{-1}(1 - x)⁠. (2 marks)

Exam answer

(i) On [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]⁠, sine is one-one and takes every value in [−1,1][-1, 1]⁠.

Why?

Sine has an inverse on an interval A only if, on A, it is one-one and takes every value from −1-1 to 1 exactly once. Any stretch of the graph that goes once from a peak to a trough, or from a trough to a peak, will do.

On [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right] the graph falls steadily from sin⁡π2=1\sin\frac{\pi}{2} = 1 to sin⁡3π2=−1\sin\frac{3\pi}{2} = -1⁠.

One such interval is A=[π2,3π2]A = \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]⁠.

Why?

It is not the principal value branch [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, as the question asks. Another correct answer is [−3π2,−π2]\left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right]⁠.

(ii) The principal value branch is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

sin⁡−1(−12)−sin⁡−1(1)=−π6−π2=−4π6\begin{aligned} &\sin^{-1}\left(-\frac12\right) - \sin^{-1}(1) \\ &= -\frac{\pi}{6} - \frac{\pi}{2} = -\frac{4\pi}{6} \end{aligned}
Why?

sin⁡−1(−12)\sin^{-1}(-\frac12) is the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] whose sine is −12-\frac12⁠. Since sin⁡(−π6)=−12\sin\left(-\frac{\pi}{6}\right) = -\frac12⁠, it is −π6-\frac{\pi}{6}⁠.

sin⁡(−π6)=−sin⁡π6=−12\begin{aligned} \sin\Big(-\frac{\pi}{6}\Big) &= -\sin\frac{\pi}{6} \\ &= -\frac12 \end{aligned}

(Sine is odd: sin⁡(−α)=−sin⁡α\sin(-\alpha) = -\sin\alpha⁠. And −π6-\frac{\pi}{6} lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.)

Similarly sin⁡π2=1\sin\frac{\pi}{2} = 1⁠, so sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2}⁠.

Watch out: 7π6\frac{7\pi}{6} also has sine −12-\frac12⁠, but it lies outside the principal branch.

=−2π3= -\frac{2\pi}{3}

Why?

Over a common denominator, −π6−3π6=−4π6-\frac{\pi}{6} - \frac{3\pi}{6} = -\frac{4\pi}{6}⁠, which simplifies to −2π3-\frac{2\pi}{3}⁠.

π2=π×32×3=3π6−π6−3π6=−π−3π6=−4π6−4π6=−4π÷26÷2=−2π3\begin{aligned} \frac{\pi}{2} &= \frac{\pi \times 3}{2 \times 3} = \frac{3\pi}{6} \\[4pt] -\frac{\pi}{6} - \frac{3\pi}{6} &= \frac{-\pi - 3\pi}{6} = -\frac{4\pi}{6} \\[4pt] -\frac{4\pi}{6} &= -\frac{4\pi \div 2}{6 \div 2} = -\frac{2\pi}{3} \end{aligned}

(iii) (a) y=sin⁡−1xy = \sin^{-1}x has domain [−1,1][-1, 1] and range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

Why?

An inverse function swaps the roles of input and output. So the domain and range of sine on its principal branch change places.

It is increasing and passes through (−1,−π2)\left(-1, -\frac{\pi}{2}\right)⁠, (0,0)(0, 0) and (1,π2)\left(1, \frac{\pi}{2}\right)⁠.

Why?

Each point (a,b)(a, b) of y=sin⁡xy = \sin x on [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] gives the point (b,a)(b, a) of y=sin⁡−1xy = \sin^{-1}x⁠. For example, (π2,1)\left(\frac{\pi}{2}, 1\right) becomes (1,π2)\left(1, \frac{\pi}{2}\right)⁠.

So the graph is the mirror image of that piece of the sine curve in the line y=xy = x⁠. It rises because sine rises on the principal branch, and it becomes vertical at the two ends.

π/2 −π/2 1 −1 O X Y
The graph of y=sin⁡−1xy = \sin^{-1}x⁠, −1≤x≤1-1 \le x \le 1⁠, with values from −π2-\frac{\pi}{2} to π2\frac{\pi}{2}⁠.

(iii) (b) f(x)=2sin⁡−1(1−x)f(x) = 2\sin^{-1}(1 - x) is defined when

−1≤1−x≤1⇒ −2≤−x≤0⇒ 0≤x≤2\begin{aligned} &-1 \le 1 - x \le 1 \\ \Rightarrow\ &-2 \le -x \le 0 \\ \Rightarrow\ &0 \le x \le 2 \end{aligned}
Why?

sin⁡−1u\sin^{-1}u is defined only for −1≤u≤1-1 \le u \le 1⁠, and here u=1−xu = 1 - x⁠.

Subtract 1 from all three parts. Then multiply by −1-1⁠, which reverses both inequality signs: −2≤−x≤0-2 \le -x \le 0 becomes 2≥x≥02 \ge x \ge 0⁠.

−1≤1−x≤1−1−1≤1−x−1≤1−1−2≤−x≤0(−1)(−2)≥(−1)(−x)≥(−1)(0)2≥x≥0\begin{aligned} -1 &\le 1 - x \le 1 \\ -1 - 1 &\le 1 - x - 1 \le 1 - 1 \\ -2 &\le -x \le 0 \\ (-1)(-2) &\ge (-1)(-x) \ge (-1)(0) \\ 2 &\ge x \ge 0 \end{aligned}

Read from the small end: 0≤x≤20 \le x \le 2⁠.

Watch out: forgetting to reverse the signs gives a wrong interval.

Domain =[0,2]= [0, 2]

−π2≤sin⁡−1(1−x)≤π2⇒ −π≤2sin⁡−1(1−x)≤π\begin{aligned} &-\frac{\pi}{2} \le \sin^{-1}(1 - x) \le \frac{\pi}{2} \\ \Rightarrow\ &-\pi \le 2\sin^{-1}(1 - x) \le \pi \end{aligned}
Why?

As xx runs over [0,2][0, 2]⁠, 1−x1 - x runs over all of [−1,1][-1, 1]⁠. So sin⁡−1(1−x)\sin^{-1}(1 - x) takes every value of the principal branch [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

Multiplying by 2, a positive number, keeps the inequality signs and doubles the ends.

2(−π2)=−π2(π2)=π\begin{aligned} 2\Big(-\frac{\pi}{2}\Big) &= -\pi \\ 2\Big(\frac{\pi}{2}\Big) &= \pi \end{aligned}

∴\therefore Domain =[0,2]= [0, 2] and range =[−π,π]= [-\pi, \pi]⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Inverse sine: branches, value, graph, domain.

Question type: (ii) Adding principal values + A negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1} (combined)

How to spot it: click to see

(ii) Type 4. Adding principal values: one at a time, then add

Spot it: two or three inverses joined by plus or minus signs, and the question asks for the value.

Here: First, two inverses are joined by a minus sign: find sin⁡−1(−12)\sin^{-1}\left(-\frac12\right) and sin⁡−11=π2\sin^{-1}1 = \frac{\pi}{2} on their own lines, then subtract.

(ii) Type 2. A negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠: add a minus sign

Spot it: a negative number inside sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠.

Here: −12-\frac12 inside sin⁡−1\sin^{-1} is negative: sin⁡−112=π6\sin^{-1}\frac12 = \frac{\pi}{6}⁠, so sin⁡−1(−12)=−π6\sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 1, Relations and Functions.

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Inverse Trigonometric Functions in past papers: question 5 of 18 · see them all