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Chapter 2 · Inverse Trigonometric Functions

Exercise 2.2 · Q1: 3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1}x=\sin^{-1}(3x-4x^3)

Exercise 2.2Properties of inverse trigonometric functions · Section 2.3
The question

Prove the following:

3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12].3\sin^{-1}x = \sin^{-1}\left(3x - 4x^3\right), \qquad x \in \left[-\frac12, \frac12\right].
Exam answer

Let x=sin⁡θx = \sin\theta⁠, where θ=sin⁡−1x\theta = \sin^{-1}x⁠.

As −12≤x≤12-\frac12 \le x \le \frac12⁠, −π6≤θ≤π6-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}⁠.

Using sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠,

RHS=sin⁡−1(3x−4x3)=sin⁡−1(3sin⁡θ−4sin⁡3θ)=sin⁡−1(sin⁡3θ)=3θ[as −π2≤3θ≤π2]\begin{aligned} \text{RHS} &= \sin^{-1}\left(3x - 4x^3\right) \\ &= \sin^{-1}\left(3\sin\theta - 4\sin^3\theta\right) \\ &= \sin^{-1}(\sin 3\theta) \\ &= 3\theta \quad[\text{as } {-\tfrac{\pi}{2}} \le 3\theta \le \tfrac{\pi}{2}] \end{aligned}

∴ RHS=3sin⁡−1x=LHS\therefore\ \text{RHS} = 3\sin^{-1}x = \text{LHS}⁠. Hence proved.

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

The one idea this problem is built on

The expression 3x−4x33x-4x^3 is the right-hand side of the triple-angle formula sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠. So if we write x=sin⁡θx = \sin\theta⁠, the right side becomes sin⁡−1(sin⁡3θ)\sin^{-1}(\sin 3\theta)⁠. That equals 3θ3\theta only when 3θ3\theta lies in the principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. The condition −12≤x≤12-\frac12 \le x \le \frac12 is exactly what keeps 3θ3\theta there.

Given, to show, and the plan
Given
A real number xx with −12≤x≤12-\frac12 \le x \le \frac12⁠.
To show
3sin⁡−1x=sin⁡−1(3x−4x3)3\sin^{-1}x = \sin^{-1}(3x-4x^3)⁠.
Plan
(1) Put θ=sin⁡−1x\theta = \sin^{-1}x⁠, so x=sin⁡θx = \sin\theta⁠. (2) Derive sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠. (3) Rewrite the right side as sin⁡−1(sin⁡3θ)\sin^{-1}(\sin 3\theta)⁠. (4) Show that 3θ3\theta lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta⁠. (5) See what goes wrong outside the interval.

Basics you need

1The principal branch of sin⁡−1\sin^{-1}

sin⁡−1\sin^{-1} has domain [−1,1][-1,1] and range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. For y=sin⁡−1xy = \sin^{-1}x⁠, yy is the one angle in that range whose sine is xx⁠. Basics: domain and range of sin and cos

2Undoing works only inside the range

sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x for every x∈[−1,1]x \in [-1,1]⁠. But sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y) = y only when y∈[−π2,π2]y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. For example sin⁡−1(sin⁡π)=sin⁡−10=0\sin^{-1}(\sin \pi) = \sin^{-1}0 = 0⁠, not π\pi⁠. Basics: what an inverse needs

3Compound and double angles

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B⁠, sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta⁠, cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta⁠, and cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta⁠. These come from Class XI trigonometry. Basics: sin and cos on the unit circle

4sin⁡−1\sin^{-1} is increasing

On [−1,1][-1,1]⁠, a bigger input gives a bigger output. So inequalities between xx-values carry over to inequalities between the angles sin⁡−1x\sin^{-1}x⁠, in the same direction.

Solution

  1. Substitute x=sin⁡θx = \sin\theta

    Let θ=sin⁡−1x\theta = \sin^{-1}x⁠. Because x∈[−12,12]⊂[−1,1]x \in \left[-\frac12, \frac12\right] \subset [-1,1]⁠, this is defined, and by the definition of sin⁡−1\sin^{-1}

    sin⁡θ=x,θ∈[−π2,π2].\sin\theta = x, \qquad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

    The left side of the identity is now simply 3sin⁡−1x=3θ3\sin^{-1}x = 3\theta⁠.

    We want to use a trigonometric identity, so we need xx to be the sine of some angle. Taking that angle to be θ=sin⁡−1x\theta = \sin^{-1}x (and not some other angle with sine xx⁠) is the key choice: it ties θ\theta to the left side and tells us exactly which interval θ\theta lives in.
    sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x for all x∈[−1,1]x\in[-1,1] (Basic 2). Lesson: substitutions and simplest form

  2. Derive the triple-angle formula sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta

    Write 3θ=2θ+θ3\theta = 2\theta + \theta and use the compound-angle formula:

    sin⁡3θ=sin⁡2θcos⁡θ+cos⁡2θsin⁡θ=2sin⁡θcos⁡2θ+(1−2sin⁡2θ)sin⁡θ=2sin⁡θ(1−sin⁡2θ)+sin⁡θ−2sin⁡3θ=3sin⁡θ−4sin⁡3θ.\begin{aligned} \sin 3\theta &= \sin 2\theta\cos\theta + \cos 2\theta\sin\theta \\ &= 2\sin\theta\cos^2\theta + (1-2\sin^2\theta)\sin\theta \\ &= 2\sin\theta(1-\sin^2\theta) + \sin\theta - 2\sin^3\theta \\ &= 3\sin\theta - 4\sin^3\theta. \end{aligned}
    The aim is a formula in sin⁡θ\sin\theta only, because sin⁡θ=x\sin\theta = x is what we know. So in line 2 we pick the version cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta (not cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta⁠), and in line 3 we replace cos⁡2θ\cos^2\theta by 1−sin⁡2θ1-\sin^2\theta⁠. Then everything is a polynomial in sin⁡θ\sin\theta⁠.
    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B with A=2θA = 2\theta⁠, B=θB = \theta⁠; sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta⁠; cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1-2\sin^2\theta⁠; sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 (Basic 3).

  3. Rewrite the right side

    Since x=sin⁡θx = \sin\theta⁠,

    3x−4x3=3sin⁡θ−4sin⁡3θ=sin⁡3θ,3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin 3\theta,

    so the right side of the identity is

    sin⁡−1(3x−4x3)=sin⁡−1(sin⁡3θ).\sin^{-1}\left(3x - 4x^3\right) = \sin^{-1}(\sin 3\theta).

    This also shows the right side is always defined: 3x−4x33x - 4x^3 is a sine, so it lies in [−1,1][-1,1]⁠.

    This is the moment where the substitution pays off: a cubic in xx collapses into a single sine. What remains is the question every inverse-trig problem comes down to: does sin⁡−1\sin^{-1} undo this sin⁡\sin⁠?

  4. Show 3θ3\theta lies in the principal range

    We have −12≤x≤12-\frac12 \le x \le \frac12⁠. Apply the increasing function sin⁡−1\sin^{-1} to all three parts (Basic 4):

    sin⁡−1(−12)≤sin⁡−1x≤sin⁡−112,\sin^{-1}\left(-\tfrac12\right) \le \sin^{-1}x \le \sin^{-1}\tfrac12,

    that is,

    −π6≤θ≤π6.-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}.

    Multiply by 3:

    −π2≤3θ≤π2.-\frac{\pi}{2} \le 3\theta \le \frac{\pi}{2}.

    So 3θ3\theta is in the principal range of sin⁡−1\sin^{-1}⁠, and therefore sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta⁠.

    sin⁡−1(sin⁡y)\sin^{-1}(\sin y) returns the angle in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with the same sine as yy⁠. When yy is already in that interval, the answer is yy itself. We used sin⁡−112=π6\sin^{-1}\tfrac12 = \frac{\pi}{6} because sin⁡π6=12\sin\frac{\pi}{6} = \frac12 and π6\frac{\pi}{6} is in the principal range.
    sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y) = y for y∈[−π2,π2]y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Section 2.3). Lesson: sin⁻¹(sin x) and the folding rules

  5. Conclude

    sin⁡−1(3x−4x3)=sin⁡−1(sin⁡3θ)=3θ=3sin⁡−1x.\sin^{-1}\left(3x-4x^3\right) = \sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x.

    This holds for every x∈[−12,12]x \in \left[-\frac12, \frac12\right]⁠, which is what was to be proved.

  6. What goes wrong outside [−12,12]\left[-\frac12, \frac12\right]

    Take 12<x≤1\frac12 \lt x \le 1⁠. Then π6<θ≤π2\frac{\pi}{6} \lt \theta \le \frac{\pi}{2}⁠, so

    π2<3θ≤3π2,\frac{\pi}{2} \lt 3\theta \le \frac{3\pi}{2},

    which is outside the principal range. The angle π−3θ\pi - 3\theta has the same sine (sin⁡(π−y)=sin⁡y\sin(\pi - y) = \sin y⁠) and lies in [−π2,π2)\left[-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, so

    sin⁡−1(3x−4x3)=π−3sin⁡−1xfor 12<x≤1.\sin^{-1}(3x-4x^3) = \pi - 3\sin^{-1}x \quad\text{for } \tfrac12 \lt x \le 1.

    This equals 3sin⁡−1x3\sin^{-1}x only if 3θ=π23\theta = \frac{\pi}{2}⁠, that is x=12x = \frac12⁠, which is excluded. So the identity fails at every xx in (12,1]\left(\frac12, 1\right]⁠. For example:

    • x=1x = 1⁠: left side 3⋅π2=3π23\cdot\frac{\pi}{2} = \frac{3\pi}{2}⁠; right side sin⁡−1(3−4)=sin⁡−1(−1)=−π2\sin^{-1}(3-4) = \sin^{-1}(-1) = -\frac{\pi}{2}⁠.
    • x=12x = \frac{1}{\sqrt2}⁠: 3x−4x3=32−422=123x - 4x^3 = \frac{3}{\sqrt2} - \frac{4}{2\sqrt2} = \frac{1}{\sqrt2}⁠, so the right side is π4\frac{\pi}{4}⁠, while the left side is 3⋅π4=3π43\cdot\frac{\pi}{4} = \frac{3\pi}{4}⁠.

    Both sides are odd functions of xx⁠, so the same happens on [−1,−12)\left[-1, -\frac12\right)⁠, where the right side equals −π−3sin⁡−1x-\pi - 3\sin^{-1}x⁠. The interval in the question is therefore exactly the set where the identity is true.

    −3π/2 −π/2 π/2 3π/2 −1 −1/2 1/2 1 blue: 3 sin⁻¹x gold dashed: sin⁻¹(3x − 4x³) green band: the two agree x = 1/√2: 3π/4 vs π/4
    Both sides plotted for −1≤x≤1-1 \le x \le 1 (the scales differ). The curves coincide on the green band [−12,12]\left[-\frac12, \frac12\right]⁠. Outside it, 3sin⁡−1x3\sin^{-1}x goes above π2\frac{\pi}{2} (or below −π2-\frac{\pi}{2}⁠), but sin⁡−1\sin^{-1} never outputs anything outside [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so the right side turns back. The red dots mark x=12x = \frac{1}{\sqrt2}⁠.
    Algebra alone cannot see this: the line 3x−4x3=sin⁡3θ3x-4x^3 = \sin 3\theta is true for every xx⁠. The failure comes entirely from the last step, sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta⁠, which needs 3θ3\theta in the principal range. That is why the question states an interval.
    sin⁡(π−y)=sin⁡y\sin(\pi - y) = \sin y⁠: the supplementary angle has the same sine. Basics: special angles and reference angles
Result

With θ=sin⁡−1x\theta = \sin^{-1}x⁠: 3x−4x3=sin⁡3θ3x - 4x^3 = \sin 3\theta⁠, and −12≤x≤12-\frac12 \le x \le \frac12 gives −π2≤3θ≤π2-\frac{\pi}{2} \le 3\theta \le \frac{\pi}{2}⁠. Hence sin⁡−1(3x−4x3)=3θ=3sin⁡−1x\sin^{-1}(3x - 4x^3) = 3\theta = 3\sin^{-1}x on [−12,12]\left[-\frac12, \frac12\right]⁠. (A proof question; the answer key gives no answer.) For 12<∣x∣≤1\frac12 \lt |x| \le 1 the identity is false.

Check it with numbers

Inside, x=0.3x = 0.3⁠: sin⁡−10.3=0.304693\sin^{-1}0.3 = 0.304693⁠, so the left side is 0.9140780.914078⁠. Also 3(0.3)−4(0.3)33(0.3) - 4(0.3)^3 =0.9−0.108=0.792= 0.9 - 0.108 = 0.792⁠, and sin⁡−10.792=0.914078\sin^{-1}0.792 = 0.914078⁠. ✓ Equal.

Outside, x=0.8x = 0.8⁠: sin⁡−10.8=0.927295\sin^{-1}0.8 = 0.927295⁠, left side 2.7818862.781886⁠. But 3(0.8)−4(0.8)33(0.8) - 4(0.8)^3 =2.4−2.048=0.352= 2.4 - 2.048 = 0.352 and sin⁡−10.352=0.359707\sin^{-1}0.352 = 0.359707⁠, which is π−2.781886\pi - 2.781886⁠, exactly as Step 6 predicts.

Common mistakes
  1. Writing sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta without checking. This is the whole content of the question. Without the range check the "proof" would also "prove" the false statement at x=1x = 1⁠.
  2. Getting the range of θ\theta wrong. x≤12x \le \frac12 gives θ≤π6\theta \le \frac{\pi}{6}⁠, not θ≤π2\theta \le \frac{\pi}{2}⁠. Use the actual interval of xx⁠, not just the domain of sin⁡−1\sin^{-1}⁠.
  3. Mixing up the triple-angle formulas. sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠, but cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta (Q2). If unsure, re-derive as in Step 2.
What to take away
  1. Recognise the pattern: 3x−4x33x - 4x^3 is sin⁡3θ\sin 3\theta in disguise when x=sin⁡θx = \sin\theta⁠.
  2. Every identity of the form sin⁡−1(sin⁡(… ))=…\sin^{-1}(\sin(\dots)) = \dots needs a range check, and the stated interval of xx is what makes it pass.
  3. Outside the interval the identity is replaced by a "folded" version, here π−3sin⁡−1x\pi - 3\sin^{-1}x⁠.

Every fact used, and where it comes from

FactUsed inSource
Principal branch of sin⁡−1\sin^{-1}⁠: domain [−1,1][-1,1]⁠, range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Steps 1, 4, 6Section 2.2, table of principal value branches (Lesson: principal values)
sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x) = x⁠; sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y) = y for y∈[−π2,π2]y\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Steps 1, 4Section 2.3
sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\thetaSteps 2, 3Derived in Step 2 from Class XI compound-angle formulas
sin⁡−112=π6\sin^{-1}\frac12 = \frac{\pi}{6}⁠; sin⁡−1\sin^{-1} is increasingStep 4sin⁡π6=12\sin\frac{\pi}{6} = \frac12⁠; graph of sin⁡−1\sin^{-1}⁠, Section 2.2
sin⁡(π−y)=sin⁡y\sin(\pi - y) = \sin yStep 6Class XI trigonometry
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