Chapter 2 · Inverse Trigonometric Functions

Exercise 2.1 · Q12: Value of $\cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right)$

Exercise 2.1Principal values · Section 2.2

The question

Find the value of $\cos^{-1}\left(\dfrac12\right) + 2\sin^{-1}\left(\dfrac12\right)$ ⁠.

Exam answer

$\cos^{-1}\left(\frac12\right) = \frac{\pi}{3}$ ⁠, as $\cos\frac{\pi}{3} = \frac12$ and $\frac{\pi}{3} \in [0, \pi]$ ⁠.

$\sin^{-1}\left(\frac12\right) = \frac{\pi}{6}$ ⁠, as $\sin\frac{\pi}{6} = \frac12$ and $\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠.

\begin{aligned} &\cos^{-1}\left(\frac12\right) + 2\sin^{-1}\left(\frac12\right) \\ &= \frac{\pi}{3} + 2\cdot\frac{\pi}{6} \end{aligned}

$\therefore$ The required value is $\dfrac{2\pi}{3}$ ⁠.

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

The one idea this problem is built on

$\cos^{-1}\left(\frac12\right)$ and $\sin^{-1}\left(\frac12\right)$ are two definite angles: $\frac{\pi}{3}$ and $\frac{\pi}{6}$ ⁠. The "2" in front of $\sin^{-1}$ simply doubles the angle after it has been found, so the expression is $\frac{\pi}{3} + 2\cdot\frac{\pi}{6}$ ⁠. It does not mean $\sin^{-1}$ of $2\cdot\frac12$ ⁠.

Given, to find, and the plan

Given: $\cos^{-1}\left(\frac12\right)$ and $\sin^{-1}\left(\frac12\right)$ ⁠; the input $\frac12$ lies in $[-1,1]$ ⁠, the domain of both.
To find: The value of $\cos^{-1}\left(\frac12\right) + 2\sin^{-1}\left(\frac12\right)$ ⁠.
Plan: (1) Find $\cos^{-1}\left(\frac12\right)$ in $[0,\pi]$ ⁠. (2) Find $\sin^{-1}\left(\frac12\right)$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ⁠. (3) Double the second value and add, with exact fractions of $\pi$ ⁠.

Basics you need

1Two principal ranges

$\cos^{-1} : [-1,1] \to [0,\pi]$ and $\sin^{-1} : [-1,1] \to \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ⁠. Lesson: principal values

2Special angles

$\cos\frac{\pi}{3} = \frac12$ and $\sin\frac{\pi}{6} = \frac12$ ⁠. (Also $\sin\frac{\pi}{3} = \frac{\sqrt3}{2}$ and $\cos\frac{\pi}{6} = \frac{\sqrt3}{2}$ ⁠.) Basics: special angles and reference angles

3What $2\sin^{-1}x$ means

$2\sin^{-1}x$ is $2 \times (\sin^{-1}x)$ ⁠: first find the angle, then multiply it by 2. It is not $\sin^{-1}(2x)$ ⁠. Basics: what an inverse needs

4cos and sin on the unit circle

$\cos y$ is the $x$ -coordinate and $\sin y$ the $y$ -coordinate of the point at angle $y$ ⁠. Basics: sin and cos on the unit circle

Solution

First term: $\cos^{-1}\left(\frac12\right)$

Let $a = \cos^{-1}\left(\frac12\right)$ ⁠. Then $\cos a = \frac12$ and $0 \le a \le \pi$ ⁠.

Reference angle: $\cos\frac{\pi}{3} = \frac12$ ⁠, so $\alpha = \frac{\pi}{3}$ ⁠. Sign: $\frac12 > 0$ ⁠, and on $[0,\pi]$ cos is positive exactly on $\left[0,\frac{\pi}{2}\right)$ ⁠, so $a$ is the acute angle itself:
$a = \cos^{-1}\left(\frac12\right) = \frac{\pi}{3}.$
Exclusion: all solutions of $\cos a = \frac12$ are $\pm\frac{\pi}{3} + 2n\pi$ ⁠. The nearest others, $-\frac{\pi}{3}$ and $\frac{5\pi}{3}$ ⁠, are outside $[0,\pi]$ ⁠.

For a positive input the principal value of $\cos^{-1}$ is the acute reference angle. The other angle on the same vertical line $x = \frac12$ ⁠, namely $-\frac{\pi}{3}$ ⁠, is in the lower half of the circle, outside the range (left panel of the figure).

$\cos^{-1} : [-1,1] \to [0,\pi]$ (Section 2.2). Basics: domain and range of sin and cos
Second term: $\sin^{-1}\left(\frac12\right)$

Let $b = \sin^{-1}\left(\frac12\right)$ ⁠. Then $\sin b = \frac12$ and $-\frac{\pi}{2} \le b \le \frac{\pi}{2}$ ⁠.

Reference angle: $\sin\frac{\pi}{6} = \frac12$ ⁠, so $\alpha = \frac{\pi}{6}$ ⁠. Sign: $\frac12 > 0$ ⁠, and on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ sin is positive exactly on $\left(0,\frac{\pi}{2}\right]$ ⁠, so
$b = \sin^{-1}\left(\frac12\right) = \frac{\pi}{6}.$
Exclusion: all solutions of $\sin b = \frac12$ are $\frac{\pi}{6} + 2n\pi$ and $\frac{5\pi}{6} + 2n\pi$ ⁠. The nearest others, $\frac{5\pi}{6}$ and $-\frac{7\pi}{6}$ ⁠, are outside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ⁠.

The point at $\frac{5\pi}{6}$ has the same height $\frac12$ as the point at $\frac{\pi}{6}$ ⁠, but it is on the left half of the circle, outside the range of $\sin^{-1}$ (right panel of the figure).

$\sin^{-1} : [-1,1] \to \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Section 2.2). Lesson: principal values
Double the second value and add

The coefficient 2 multiplies the angle $b$ ⁠:
$2\sin^{-1}\left(\frac12\right) = 2\cdot\frac{\pi}{6} = \frac{\pi}{3}.$
So the whole expression is
$\begin{aligned} a + 2b &= \frac{\pi}{3} + \frac{\pi}{3}\\ &= \frac{2\pi}{3}. \end{aligned}$

Left: $\cos^{-1}\left(\frac12\right) = \frac{\pi}{3}$ ⁠, the point of the upper half (green) on the line $x = \frac12$ ⁠. Right: $\sin^{-1}\left(\frac12\right) = \frac{\pi}{6}$ ⁠, the point of the right half (green) on the line $y = \frac12$ ⁠. Rose points have the same ratio but lie outside the range.

Once each inverse trig term has been replaced by its value, the rest is ordinary arithmetic with fractions of $\pi$ ⁠. The result $\frac{2\pi}{3}$ does not have to lie in any principal range: it is a sum of two angles, not the output of an inverse function.

$2\sin^{-1}x$ means $2 \times \sin^{-1}x$ (Basic 3). Basics: angles and radians

Result

\cos^{-1}\left(\frac12\right) + 2\sin^{-1}\left(\frac12\right) = \frac{2\pi}{3}.

The terms are $\frac{\pi}{3}$ and $2\cdot\frac{\pi}{6} = \frac{\pi}{3}$ ⁠. (Textbook answer: $\frac{2\pi}{3}$ ⁠.)

Check it with numbers

In radians: $\cos^{-1}(0.5) \approx 1.0471976$ and $\sin^{-1}(0.5) \approx 0.5235988$ ⁠. Then $1.0471976 + 2 \times 0.5235988$ is $\approx 2.0943951$ ⁠, and $\frac{2\pi}{3} \approx 2.0943951$ ✓.

Common mistakes

Reading $2\sin^{-1}\left(\frac12\right)$ as $\sin^{-1}\left(2\cdot\frac12\right) = \sin^{-1}(1) = \frac{\pi}{2}$ ⁠. This gives the wrong total $\frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$ ⁠. The 2 multiplies the angle, not the input.
Swapping the special angles. $\cos^{-1}\left(\frac12\right) = \frac{\pi}{3}$ but $\sin^{-1}\left(\frac12\right) = \frac{\pi}{6}$ ⁠. Swapping gives $\frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6}$ ⁠, also wrong.
Using $\frac{5\pi}{6}$ for $\sin^{-1}\left(\frac12\right)$ ⁠. It has sine $\frac12$ ⁠, but it is outside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ ⁠.

What to take away

Replace each inverse trig term by its principal value first; coefficients then act on those angles.
$\sin^{-1}x$ and $\cos^{-1}x$ of the same $x$ are complementary here: $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$ ⁠. That is why $2\sin^{-1}\left(\frac12\right)$ happens to equal $\cos^{-1}\left(\frac12\right)$ ⁠.
The final sum need not lie in any principal range.

Every fact used, and where it comes from

Fact	Used in	Source
Principal ranges of $\cos^{-1}$ and $\sin^{-1}$	Steps 1, 2	Section 2.2 and its table of principal value branches (Lesson: principal values)
$\cos\frac{\pi}{3} = \frac12$ ⁠, $\sin\frac{\pi}{6} = \frac12$	Steps 1, 2	Class XI trigonometry (special angles)
Periods of sin and cos; $\sin(\pi - \theta) = \sin\theta$ ⁠, $\cos(-\theta) = \cos\theta$	Steps 1, 2 (exclusion)	Class XI trigonometry (allied angles)
$\sin^{-1}$ is a function: one value per input, so $2\sin^{-1}x$ is twice that value	Step 3	Section 2.2
All numerical values	Check	Recomputed in the site's verification script

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Exercise 2.1 · Q12: Value of cos⁡−1(12)+2sin⁡−1(12)\cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right)cos−1(21​)+2sin−1(21​)

Basics you need

Solution

First term: cos⁡−1(12)\cos^{-1}\left(\frac12\right)cos−1(21​)

Second term: sin⁡−1(12)\sin^{-1}\left(\frac12\right)sin−1(21​)

Double the second value and add

Every fact used, and where it comes from

Exercise 2.1 · Q12: Value of $\cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right)$

First term: $\cos^{-1}\left(\frac12\right)$

Second term: $\sin^{-1}\left(\frac12\right)$