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Chapter 2 · Inverse Trigonometric Functions

Exercise 2.2 · Q2: 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x=\cos^{-1}(4x^3-3x)

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

Prove the following:

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

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

As 12≤x≤1\frac12 \le x \le 1⁠, 0≤θ≤π30 \le \theta \le \frac{\pi}{3}⁠.

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

RHS=cos⁡−1(4x3−3x)=cos⁡−1(4cos⁡3θ−3cos⁡θ)=cos⁡−1(cos⁡3θ)=3θ[as 0≤3θ≤π]\begin{aligned} \text{RHS} &= \cos^{-1}\left(4x^3 - 3x\right) \\ &= \cos^{-1}\left(4\cos^3\theta - 3\cos\theta\right) \\ &= \cos^{-1}(\cos 3\theta) \\ &= 3\theta \quad[\text{as } 0 \le 3\theta \le \pi] \end{aligned}

∴ RHS=3cos⁡−1x=LHS\therefore\ \text{RHS} = 3\cos^{-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

4x3−3x4x^3 - 3x is the right-hand side of cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta⁠. With x=cos⁡θx = \cos\theta⁠, the right side becomes cos⁡−1(cos⁡3θ)\cos^{-1}(\cos 3\theta)⁠, which equals 3θ3\theta only when 3θ∈[0,π]3\theta \in [0, \pi], the principal range of cos⁡−1\cos^{-1}⁠. The condition 12≤x≤1\frac12 \le x \le 1 is exactly what puts 3θ3\theta there. Notice the interval is not symmetric about 0, unlike Q1: cos⁡−1\cos^{-1} has a different range from sin⁡−1\sin^{-1}⁠.

Given, to show, and the plan
Given
A real number xx with 12≤x≤1\frac12 \le x \le 1⁠.
To show
3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)⁠.
Plan
(1) Put θ=cos⁡−1x\theta = \cos^{-1}x⁠, so x=cos⁡θx = \cos\theta⁠. (2) Derive cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta⁠. (3) Rewrite the right side as cos⁡−1(cos⁡3θ)\cos^{-1}(\cos 3\theta)⁠. (4) Show 3θ∈[0,π]3\theta \in [0,\pi]⁠. (5) See what goes wrong for x<12x \lt \frac12⁠.

Basics you need

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

cos⁡−1\cos^{-1} has domain [−1,1][-1,1] and range [0,π][0, \pi]⁠. Basics: domain and range of sin and cos

2Undoing works only inside the range

cos⁡(cos⁡−1x)=x\cos(\cos^{-1}x) = x for all x∈[−1,1]x\in[-1,1]⁠, but cos⁡−1(cos⁡y)=y\cos^{-1}(\cos y) = y only for y∈[0,π]y \in [0,\pi]⁠. For example cos⁡−1(cos⁡2π)=cos⁡−11=0\cos^{-1}(\cos 2\pi) = \cos^{-1}1 = 0⁠. Basics: what an inverse needs

3Compound and double angles

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

4cos⁡−1\cos^{-1} is decreasing

On [−1,1][-1,1]⁠, a bigger input gives a smaller angle (cos⁡−11=0\cos^{-1}1 = 0⁠, cos⁡−1(−1)=π\cos^{-1}(-1) = \pi⁠). So applying cos⁡−1\cos^{-1} to an inequality reverses it.

Solution

  1. Substitute x=cos⁡θx = \cos\theta

    Let θ=cos⁡−1x\theta = \cos^{-1}x⁠. This is defined because [12,1]⊂[−1,1]\left[\frac12, 1\right] \subset [-1,1]⁠, and

    cos⁡θ=x,θ∈[0,π].\cos\theta = x, \qquad \theta \in [0, \pi].

    The left side is 3cos⁡−1x=3θ3\cos^{-1}x = 3\theta⁠.

    Choosing θ=cos⁡−1x\theta = \cos^{-1}x itself (rather than any angle with cosine xx⁠) links θ\theta directly to the left side and pins down the interval it lies in.
    cos⁡(cos⁡−1x)=x\cos(\cos^{-1}x) = x for x∈[−1,1]x\in[-1,1]⁠. Lesson: substitutions and simplest form

  2. Derive cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta

    Write 3θ=2θ+θ3\theta = 2\theta + \theta⁠:

    cos⁡3θ=cos⁡2θcos⁡θ−sin⁡2θsin⁡θ=(2cos⁡2θ−1)cos⁡θ−2sin⁡2θcos⁡θ=2cos⁡3θ−cos⁡θ−2(1−cos⁡2θ)cos⁡θ=4cos⁡3θ−3cos⁡θ.\begin{aligned} \cos 3\theta &= \cos 2\theta\cos\theta - \sin 2\theta\sin\theta \\ &= (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta \\ &= 2\cos^3\theta - \cos\theta - 2(1-\cos^2\theta)\cos\theta \\ &= 4\cos^3\theta - 3\cos\theta. \end{aligned}
    We know cos⁡θ=x\cos\theta = x⁠, so we want everything in terms of cos⁡θ\cos\theta⁠. That is why we choose cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1 and replace sin⁡2θ\sin^2\theta by 1−cos⁡2θ1-\cos^2\theta⁠.
    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin B with A=2θA = 2\theta⁠, B=θB = \theta (Basic 3).

  3. Rewrite the right side

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

    so

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

    Since 4x3−3x4x^3 - 3x is a cosine, it is always in [−1,1][-1,1] and the right side is defined.

  4. Show 3θ3\theta lies in [0,π][0, \pi]

    From 12≤x≤1\frac12 \le x \le 1⁠, apply the decreasing function cos⁡−1\cos^{-1}⁠; the inequalities reverse (Basic 4):

    cos⁡−11≤cos⁡−1x≤cos⁡−112,\cos^{-1}1 \le \cos^{-1}x \le \cos^{-1}\tfrac12,

    that is,

    0≤θ≤π3.0 \le \theta \le \frac{\pi}{3}.

    Multiply by 3:

    0≤3θ≤π.0 \le 3\theta \le \pi.

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

    The largest xx (namely 1) gives the smallest angle (0), and the smallest xx (namely 12\frac12⁠) gives the largest angle π3\frac{\pi}{3}⁠, because cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠. Tripling π3\frac{\pi}{3} gives exactly π\pi⁠, the top of the principal range.
    cos⁡−1(cos⁡y)=y\cos^{-1}(\cos y) = y for y∈[0,π]y\in[0,\pi] (Section 2.3). Lesson: sin⁻¹(sin x) and the folding rules

  5. Conclude

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

    for every x∈[12,1]x\in\left[\frac12, 1\right]⁠.

  6. What goes wrong for x<12x \lt \frac12

    If −12≤x<12-\frac12 \le x \lt \frac12⁠, then π3<θ≤2π3\frac{\pi}{3} \lt \theta \le \frac{2\pi}{3}⁠, so π<3θ≤2π\pi \lt 3\theta \le 2\pi⁠: outside [0,π][0,\pi]⁠. The angle 2π−3θ2\pi - 3\theta has the same cosine and lies in [0,π)[0, \pi)⁠, so

    cos⁡−1(4x3−3x)=2π−3cos⁡−1x.\cos^{-1}(4x^3 - 3x) = 2\pi - 3\cos^{-1}x.

    If −1≤x<−12-1 \le x \lt -\frac12⁠, then 2π<3θ≤3π2\pi \lt 3\theta \le 3\pi⁠, and 3θ−2π∈(0,π]3\theta - 2\pi \in (0, \pi]⁠, so

    cos⁡−1(4x3−3x)=3cos⁡−1x−2π.\cos^{-1}(4x^3 - 3x) = 3\cos^{-1}x - 2\pi.

    Neither equals 3θ3\theta⁠: the first would need θ=π3\theta = \frac{\pi}{3} (that is x=12x = \frac12⁠, excluded), the second would need 2π=02\pi = 0⁠. So the identity fails at every x∈[−1,12)x \in \left[-1, \frac12\right)⁠. For example:

    • x=0x = 0⁠: left side 3⋅π2=3π23\cdot\frac{\pi}{2} = \frac{3\pi}{2}⁠; right side cos⁡−10=π2\cos^{-1}0 = \frac{\pi}{2}⁠.
    • x=−1x = -1⁠: left side 3π3\pi⁠; right side cos⁡−1(−4+3)=cos⁡−1(−1)=π\cos^{-1}(-4+3) = \cos^{-1}(-1) = \pi⁠.
    π/2 π 3π/2 2π 5π/2 3π −1 −1/2 1/2 1 blue: 3 cos⁻¹x gold dashed: cos⁻¹(4x³ − 3x) green band: they agree x = 0: 3π/2 vs π/2
    Both sides for −1≤x≤1-1 \le x \le 1 (the scales differ). They coincide only on the green band [12,1]\left[\frac12, 1\right]⁠. Left of it, 3cos⁡−1x3\cos^{-1}x rises above π\pi⁠, but cos⁡−1\cos^{-1} never outputs more than π\pi⁠, so the right side folds back down. The red dots mark x=0x = 0⁠.
    The algebraic identity 4x3−3x=cos⁡3θ4x^3 - 3x = \cos 3\theta is true for every xx⁠; the failure comes only from cos⁡−1(cos⁡3θ)=3θ\cos^{-1}(\cos 3\theta) = 3\theta⁠, which needs 3θ∈[0,π]3\theta\in[0,\pi]⁠. So the stated interval is exactly the set where the identity holds.
    cos⁡(2π−y)=cos⁡y\cos(2\pi - y) = \cos y and cos⁡(y−2π)=cos⁡y\cos(y - 2\pi) = \cos y⁠: cosine has period 2π2\pi and is even. Basics: special angles and reference angles
Result

With θ=cos⁡−1x\theta = \cos^{-1}x⁠: 4x3−3x=cos⁡3θ4x^3 - 3x = \cos 3\theta⁠, and 12≤x≤1\frac12 \le x \le 1 gives 0≤3θ≤π0 \le 3\theta \le \pi⁠. Hence cos⁡−1(4x3−3x)=3θ=3cos⁡−1x\cos^{-1}(4x^3 - 3x) = 3\theta = 3\cos^{-1}x on [12,1]\left[\frac12, 1\right]⁠. (A proof question; the answer key gives no answer.) For −1≤x<12-1 \le x \lt \frac12 the identity is false.

Check it with numbers

Inside, x=0.7x = 0.7⁠: cos⁡−10.7=0.795399\cos^{-1}0.7 = 0.795399⁠, left side 2.3861962.386196⁠. And 4(0.7)3−3(0.7)4(0.7)^3 - 3(0.7) =1.372−2.1=−0.728= 1.372 - 2.1 = -0.728⁠, cos⁡−1(−0.728)=2.386196\cos^{-1}(-0.728) = 2.386196⁠. ✓ Equal.

Outside, x=0.2x = 0.2⁠: cos⁡−10.2=1.369438\cos^{-1}0.2 = 1.369438⁠, left side 4.1083154.108315⁠. But 4(0.2)3−3(0.2)=−0.5684(0.2)^3 - 3(0.2) = -0.568 and cos⁡−1(−0.568)=2.174870\cos^{-1}(-0.568) = 2.174870⁠, which is 2π−4.1083152\pi - 4.108315⁠, as Step 6 predicts.

Common mistakes
  1. Not reversing the inequality. cos⁡−1\cos^{-1} is decreasing, so 12≤x≤1\frac12 \le x \le 1 becomes cos⁡−11≤θ≤cos⁡−112\cos^{-1}1 \le \theta \le \cos^{-1}\frac12⁠.
  2. Copying the interval from Q1. The range of cos⁡−1\cos^{-1} is [0,π][0,\pi]⁠, not [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so the valid interval is [12,1]\left[\frac12, 1\right]⁠, not [−12,12]\left[-\frac12, \frac12\right]⁠.
  3. Skipping the range check. Without it, the same lines would "prove" 3π=π3\pi = \pi at x=−1x = -1⁠.
What to take away
  1. 4x3−3x4x^3 - 3x is cos⁡3θ\cos 3\theta when x=cos⁡θx = \cos\theta⁠.
  2. The valid interval depends on which inverse function is used, because each has its own principal range.
  3. Outside the interval the identity changes to a folded form: 2π−3cos⁡−1x2\pi - 3\cos^{-1}x or 3cos⁡−1x−2π3\cos^{-1}x - 2\pi⁠.

Every fact used, and where it comes from

FactUsed inSource
Principal branch of cos⁡−1\cos^{-1}⁠: domain [−1,1][-1,1]⁠, range [0,π][0,\pi]Steps 1, 4, 6Section 2.2, table of principal value branches (Lesson: principal values)
cos⁡(cos⁡−1x)=x\cos(\cos^{-1}x) = x⁠; cos⁡−1(cos⁡y)=y\cos^{-1}(\cos y) = y for y∈[0,π]y\in[0,\pi]Steps 1, 4Section 2.3 ("similar results follow")
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\thetaSteps 2, 3Derived in Step 2 from Class XI compound-angle formulas
cos⁡−112=π3\cos^{-1}\frac12 = \frac{\pi}{3}⁠; cos⁡−1\cos^{-1} is decreasingStep 4cos⁡π3=12\cos\frac{\pi}{3} = \frac12⁠; graph of cos⁡−1\cos^{-1}⁠, Section 2.2
cos⁡(2π−y)=cos⁡(y−2π)=cos⁡y\cos(2\pi - y) = \cos(y - 2\pi) = \cos yStep 6Class XI trigonometry
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