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Chapter 2 · Inverse Trigonometric Functions

Exercise 2.2 · Q8: tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]

Exercise 2.2Values of nested inverse trigonometric expressions · Section 2.3
The question

Find the values of each of the following:

tan⁡−1[2cos⁡(2sin⁡−112)].\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right].
Exam answer

Since sin⁡π6=12\sin\frac{\pi}{6} = \frac12 and π6∈[−π2,π2]\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, sin⁡−112=π6\sin^{-1}\frac12 = \frac{\pi}{6}⁠.

tan⁡−1[2cos⁡(2sin⁡−112)]=tan⁡−1[2cos⁡(2⋅π6)]=tan⁡−1(2cos⁡π3)=tan⁡−1(2⋅12)=tan⁡−11\begin{aligned} &\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right] \\ &= \tan^{-1}\left[2\cos\left(2\cdot\frac{\pi}{6}\right)\right] \\ &= \tan^{-1}\left(2\cos\frac{\pi}{3}\right) \\ &= \tan^{-1}\left(2\cdot\frac12\right) = \tan^{-1}1 \end{aligned}

Since tan⁡π4=1\tan\frac{\pi}{4} = 1 and π4∈(−π2,π2)\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠,

∴\therefore the required value is tan⁡−11=π4\tan^{-1}1 = \dfrac{\pi}{4}⁠.

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

The one idea this problem is built on

A nested expression is evaluated from the inside out, like peeling an onion. The innermost piece is a principal value, sin⁡−112\sin^{-1}\frac12⁠. Each layer after that is ordinary arithmetic or a familiar trigonometric value, until the last layer, tan⁡−1\tan^{-1}⁠, which is again a principal value. At each inverse-function layer we must pick the angle in the correct principal range.

Given, to find, and the plan
Given
The expression tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right]⁠.
To find
Its exact value.
Plan
Work outwards one layer at a time: (1) sin⁡−112\sin^{-1}\frac12⁠; (2) double it; (3) take the cosine; (4) double that; (5) take tan⁡−1\tan^{-1} of the result.

Basics you need

1Principal value

sin⁡−1y\sin^{-1}y is the one angle in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] whose sine is yy⁠; tan⁡−1y\tan^{-1}y is the one angle in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) whose tangent is yy⁠. Lesson: principal values

2Special angles

sin⁡π6=12\sin\frac{\pi}{6}=\frac12⁠, cos⁡π3=12\cos\frac{\pi}{3}=\frac12⁠, tan⁡π4=1\tan\frac{\pi}{4}=1⁠. Basics: special angles and reference angles

3Order of operations in a nest

In f(g(h(x)))f(g(h(x))) you compute hh first, then gg⁠, then ff⁠. Brackets tell you the order.

4Radians

π\pi radians is 180∘180^\circ⁠, so π6=30∘\frac{\pi}{6}=30^\circ⁠, π3=60∘\frac{\pi}{3}=60^\circ⁠, π4=45∘\frac{\pi}{4}=45^\circ⁠. Basics: angles and radians

Solution

  1. Innermost layer: sin⁡−112\sin^{-1}\frac12

    We need the angle in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] whose sine is 12\frac12⁠. Since sin⁡π6=12\sin\frac{\pi}{6}=\frac12 and π6\frac{\pi}{6} lies in that interval,

    sin⁡−112=π6.\sin^{-1}\frac12=\frac{\pi}{6}.
    Other angles also have sine 12\frac12⁠, for example 5π6\frac{5\pi}{6}⁠. But 5π6\frac{5\pi}{6} is outside [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, so it is not the principal value. Sine is one-one on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, so π6\frac{\pi}{6} is the only candidate there.
    This is Exercise 2.1-style work: find the angle, then confirm it is in the principal range. Basics: special angles

  2. Double it

    2sin⁡−112=2⋅π6=π3.2\sin^{-1}\frac12 = 2\cdot\frac{\pi}{6} = \frac{\pi}{3}.
    The "2" multiplies the angle, not the sine. So we double π6\frac{\pi}{6}⁠, not 12\frac12⁠.

  3. Take the cosine

    cos⁡(2sin⁡−112)=cos⁡π3=12.\cos\left(2\sin^{-1}\frac12\right)=\cos\frac{\pi}{3}=\frac12.
    This layer is an ordinary cosine of a known angle, so no range question arises: cos⁡\cos is defined for every angle and gives exactly one value.
    cos⁡60∘=12\cos60^\circ=\frac12⁠. Basics: special angles

  4. Double again

    2cos⁡(2sin⁡−112)=2⋅12=1.2\cos\left(2\sin^{-1}\frac12\right)=2\cdot\frac12=1.
    Again just arithmetic; this "2" multiplies the number 12\frac12⁠.

  5. Outer layer: tan⁡−11\tan^{-1}1

    We need the angle in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) whose tangent is 1. Since tan⁡π4=1\tan\frac{\pi}{4}=1 and π4\frac{\pi}{4} lies in that interval,

    tan⁡−1[2cos⁡(2sin⁡−112)]=tan⁡−11=π4.\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]=\tan^{-1}1=\frac{\pi}{4}.
    5π4\frac{5\pi}{4} also has tangent 1, but it lies outside the principal range (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)⁠. Tangent is one-one on that interval, so π4\frac{\pi}{4} is the only answer.
    tan⁡−1:R→(−π2,π2)\tan^{-1}:\R\to\left(-\frac{\pi}{2},\frac{\pi}{2}\right) (table of principal value branches, Section 2.2). Basics: tan, cot, sec and cosec
Result

tan⁡−1[2cos⁡(2sin⁡−112)]=π4.\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]=\frac{\pi}{4}. (Textbook answer: π4\frac{\pi}{4}⁠.)

Check it with numbers

On a calculator in radian mode: sin⁡−10.5=0.5235987756\sin^{-1}0.5=0.5235987756⁠; doubled, 1.04719755121.0471975512⁠; its cosine is 0.50.5⁠; doubled, 11⁠; and tan⁡−11=0.7853981634\tan^{-1}1=0.7853981634⁠, which is π4\frac{\pi}{4} (since π≈3.1415926536\pi\approx3.1415926536⁠). ✓

Common mistakes
  1. Working from the outside in. You cannot take tan⁡−1\tan^{-1} of something you have not computed yet. Start with the innermost bracket.
  2. Doubling the wrong thing. 2sin⁡−1122\sin^{-1}\frac12 is twice the angle π6\frac{\pi}{6}⁠; it is not sin⁡−11\sin^{-1}1⁠.
  3. Choosing a non-principal angle. sin⁡−112\sin^{-1}\frac12 is not 5π6\frac{5\pi}{6}⁠, and tan⁡−11\tan^{-1}1 is not 5π4\frac{5\pi}{4}⁠.
  4. Calculator in degree mode. Then sin⁡−10.5\sin^{-1}0.5 shows 3030⁠; the final answer 45∘45^\circ is the same angle as π4\frac{\pi}{4}⁠, but write it in radians.
What to take away
  1. Nested expressions are peeled from the inside out.
  2. Every sin⁡−1\sin^{-1}⁠, cos⁡−1\cos^{-1}⁠, tan⁡−1\tan^{-1} layer returns a principal value; every sin⁡\sin⁠, cos⁡\cos⁠, tan⁡\tan layer is a plain evaluation.
  3. Know the special-angle table by heart; these problems are built from it.

Every fact used, and where it comes from

FactUsed inSource
Principal ranges: sin⁡−1\sin^{-1} in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]⁠, tan⁡−1\tan^{-1} in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)Steps 1, 5Table of principal value branches, Section 2.2
sin⁡π6=12\sin\frac{\pi}{6}=\frac12⁠, cos⁡π3=12\cos\frac{\pi}{3}=\frac12⁠, tan⁡π4=1\tan\frac{\pi}{4}=1Steps 1, 3, 5Special angles, Class XI, Chapter 3
Principal value of sin⁡−112\sin^{-1}\frac{1}{\sqrt2} found the same wayStep 1 (method)Example 1, Section 2.2
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