INVERSE TRIGONOMETRIC FUNCTIONS

Chapter 2

Inverse Trigonometric Functions

Big idea: sine, cosine and tangent turn an angle into a number. Their inverses turn a number back into an angle. Because many angles share the same sine, we must agree on one standard answer, called the principal value.

In Class XI you used trigonometric functions to go from an angle to a ratio. For example, $\sin 30° = \frac{1}{2}$. Real problems often go the other way. A ramp rises $1$ m over a flat distance of $4$ m. What is its angle with the ground? You know the ratio $\tan \theta = \frac{1}{4}$ and you want the angle $\theta$.

The tool for this is the inverse trigonometric function. Here $\theta = \tan^{-1}\!\left(\frac{1}{4}\right) \approx 14.04°$. The button on a calculator marked $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ does exactly this job.

Ask: which angle has sine equal to $\frac{1}{2}$? The angle $30°$ works, since $\sin 30° = 0.5$. But $150°$ also works: $\sin 150° = 0.5$. In fact, so do $390°$, $510°$, and infinitely many more, because the sine repeats every $360°$. A function must give one answer for each input, so we must choose. The agreed choice is a short list of angles where the function does not repeat. The answer taken from that list is called the principal value.

Angles in this chapter are measured in radians, where $\pi$ radians $= 180°$. Remember a few conversions: $\frac{\pi}{6} = 30°$, $\frac{\pi}{4} = 45°$, $\frac{\pi}{3} = 60°$, $\frac{\pi}{2} = 90°$.

A function has an inverse only if it is a bijection (Chapter 1). The sine function on all of $\mathbb{R}$ is not one-one, because it repeats. The picture shows it: the dashed line $y = \frac{1}{2}$ crosses the curve at four points between $-2\pi$ and $2\pi$, and it would keep crossing forever.

The fix is to keep one piece of the curve where it is both one-one and onto its range. On $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ the sine function rises steadily from $-1$ to $1$. So $\sin : \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to [-1, 1]$ is a bijection and has an inverse. Other pieces would also work (for example $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$), but we agree to use this one. It is called the principal value branch.

The same idea is applied to each of the six trigonometric functions. The table gives the domain (allowed inputs) and the range (the principal values that come out):

Definition. $y = \sin^{-1} x$ means $x = \sin y$ and $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. The same pattern holds for the other five, each with its own branch.

The two "undo" rules from Chapter 1 now read:

• $\sin(\sin^{-1} x) = x$ for every $x \in [-1, 1]$. This always works.
• $\sin^{-1}(\sin x) = x$ only when $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Outside this range, the answer is a different angle.

Example of the second rule failing: $\sin\frac{5\pi}{6} = \frac{1}{2}$, but $\sin^{-1}\!\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}$, not $\frac{5\pi}{6}$. The angle $\frac{5\pi}{6}$ is outside the branch, so the function returns the angle with the same sine that is inside it. Similarly $\cos^{-1}\!\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$ and $\tan^{-1}\!\left(\tan\frac{3\pi}{4}\right) = -\frac{\pi}{4}$.

The graph of an inverse function is the mirror image of the original graph in the line $y = x$ (Chapter 1). Below, the teal curve is $\sin$ on its branch and the orange curve is $\sin^{-1}$. The point $\left(\frac{\pi}{6}, \frac{1}{2}\right)$ on the first becomes $\left(\frac{1}{2}, \frac{\pi}{6}\right)$ on the second.

The graph of $\cos^{-1}$ falls from $\pi$ (at $x = -1$) to $0$ (at $x = 1$), passing through $\frac{\pi}{2}$ at $x = 0$. The graph of $\tan^{-1}$ rises for all real $x$ but never reaches $\pm\frac{\pi}{2}$: those heights are horizontal asymptotes.

These values come from the special angles $30°$, $45°$, $60°$. Each cell shows the principal value in radians, with degrees in brackets.

Find $\sin^{-1}\!\left(-\frac{1}{2}\right)$. Angles with sine $-\frac{1}{2}$ include $-30°$, $210°$ and $330°$. Only $-30°$, that is $-\frac{\pi}{6}$, lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So $\sin^{-1}\!\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Find $\cos^{-1}\!\left(-\frac{\sqrt{3}}{2}\right)$. The cosine is negative, so the angle is in $\left(\frac{\pi}{2}, \pi\right]$. Since $\cos 30° = \frac{\sqrt{3}}{2}$, the angle is $180° - 30° = 150°$. So the answer is $\frac{5\pi}{6}$.

Find $\cot^{-1}(-1)$. The branch of $\cot^{-1}$ is $(0, \pi)$, so even for a negative input the answer is positive. Since $\cot 135° = -1$, the answer is $\frac{3\pi}{4}$ (that is, $135°$). It is not $-\frac{\pi}{4}$.

Find $\cos\!\left(\sin^{-1}\frac{3}{5}\right)$. Let $\theta = \sin^{-1}\frac{3}{5}$. Draw a right triangle with opposite side $3$ and hypotenuse $5$. The adjacent side is $4$ (because $3^2 + 4^2 = 5^2$). So $\cos\theta = \frac{4}{5} = 0.8$. In the same way, $\tan\!\left(\cos^{-1}\frac{5}{13}\right) = \frac{12}{5} = 2.4$.

1. Find $\sin^{-1}(-1)$.
2. Find $\cos^{-1}\!\left(-\frac{\sqrt{2}}{2}\right)$.
3. Find $\tan^{-1}\sqrt{3}$.
4. Find $\sin^{-1}\!\left(\sin\frac{7\pi}{6}\right)$.

Answers

1. $-\frac{\pi}{2}$ (that is, $-90°$).
2. $\frac{3\pi}{4}$ (that is, $135°$).
3. $\frac{\pi}{3}$ (that is, $60°$).
4. $-\frac{\pi}{6}$. Since $\sin\frac{7\pi}{6} = -\frac{1}{2}$, the answer is the angle in the branch with sine $-\frac{1}{2}$, not $\frac{7\pi}{6}$.

1. Find the principal value of $\sin^{-1}\!\left(-\frac{\sqrt{3}}{2}\right)$.
2. Find the principal value of $\cos^{-1}\!\left(\frac{1}{\sqrt{2}}\right)$.
3. Find the principal value of $\tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right)$.
4. Find the principal value of $\cot^{-1}\sqrt{3}$.
5. Find the principal value of $\sec^{-1}(-2)$.
6. Find the principal value of $\csc^{-1}\sqrt{2}$.
7. Find the principal value of $\cos^{-1}(-1)$.
8. Find the principal value of $\sin^{-1}\!\left(\sin\frac{3\pi}{4}\right)$.
9. Find $\sin\!\left(\cos^{-1}\frac{4}{5}\right)$ and $\tan\!\left(\sin^{-1}\frac{5}{13}\right)$.
10. Is $\sin^{-1}(2)$ defined? What about $\tan^{-1}(2)$?

Answers

1. $-\frac{\pi}{3}$
2. $\frac{\pi}{4}$
3. $\frac{\pi}{6}$
4. $\frac{\pi}{6}$
5. $\frac{2\pi}{3}$
6. $\frac{\pi}{4}$
7. $\pi$
8. $\frac{\pi}{4}$
9. Use right triangles. For the first, adjacent $4$ and hypotenuse $5$ give opposite $3$, so $\sin = \frac{3}{5} = 0.6$. For the second, opposite $5$ and hypotenuse $13$ give adjacent $12$, so $\tan = \frac{5}{12} \approx 0.4167$.
10. $\sin^{-1}(2)$ is not defined: the sine of any angle is at most $1$, and the domain of $\sin^{-1}$ is $[-1, 1]$. $\tan^{-1}(2)$ is defined, because the domain of $\tan^{-1}$ is all real numbers.

In this section $x$ and $y$ are inside the domains of the functions, and the answers are principal values. Each rule below was checked by computer at many sample values, and for rules with conditions the computer also confirmed that the rule fails outside the condition.

• $\sin^{-1}(-x) = -\sin^{-1} x$ for $x \in [-1, 1]$.
• $\tan^{-1}(-x) = -\tan^{-1} x$ for all real $x$.
• $\cos^{-1}(-x) = \pi - \cos^{-1} x$ for $x \in [-1, 1]$.

Why? Sine and tangent are "odd" functions, so their inverses are odd too. Cosine is not odd: the angle for $-x$ is the mirror image of the angle for $x$ across $\frac{\pi}{2}$. Example: $\cos^{-1}\!\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\frac{1}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

• $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$ for $x \in [-1, 1]$.
• $\tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}$ for all real $x$.

Proof of the first. Let $\theta = \sin^{-1} x$, so $x = \sin\theta$. Also $\sin\theta = \cos\!\left(\frac{\pi}{2} - \theta\right)$. Because $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the angle $\frac{\pi}{2} - \theta$ is in $[0, \pi]$, which is exactly the branch of $\cos^{-1}$. So $\cos^{-1} x = \frac{\pi}{2} - \theta$, and adding gives $\frac{\pi}{2}$.

Example: $\sin^{-1} 0.3 + \cos^{-1} 0.3 = 1.5708$, which is $\frac{\pi}{2} \approx 1.5708$.

• $\sin^{-1}\dfrac{1}{x} = \csc^{-1} x$ and $\cos^{-1}\dfrac{1}{x} = \sec^{-1} x$ for $|x| \ge 1$.
• $\tan^{-1}\dfrac{1}{x} = \cot^{-1} x$ for $x > 0$. For $x < 0$ they differ by $\pi$: $\cot^{-1} x = \pi + \tan^{-1}\dfrac{1}{x}$.

Check: $\csc^{-1} 2 = \frac{\pi}{6} = \sin^{-1}\frac{1}{2}$, and $\sec^{-1}(-2) = \frac{2\pi}{3} = \cos^{-1}\!\left(-\frac{1}{2}\right)$. For a negative input, $\cot^{-1}(-1) = \frac{3\pi}{4}$, while $\tan^{-1}(-1) = -\frac{\pi}{4}$, and the difference is exactly $\pi$.

• $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$, and $\sin^{-1}(\sin x) = x$ for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
• $\cos(\cos^{-1} x) = x$ for $x \in [-1, 1]$, and $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$.
• $\tan(\tan^{-1} x) = x$ for all real $x$, and $\tan^{-1}(\tan x) = x$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

When $xy < 1$:

$$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\frac{x + y}{1 - xy}$$

When $xy > -1$:

$$\tan^{-1} x - \tan^{-1} y = \tan^{-1}\frac{x - y}{1 + xy}$$

Why. Let $A = \tan^{-1} x$ and $B = \tan^{-1} y$. The tangent addition formula gives $\tan(A + B) = \frac{x + y}{1 - xy}$. When $xy < 1$, the angle $A + B$ stays inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is the branch of $\tan^{-1}$, so taking $\tan^{-1}$ of both sides is allowed. When $xy > 1$ and both are positive, $A + B$ is bigger than $\frac{\pi}{2}$, so the formula is short by exactly $\pi$: $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\frac{x + y}{1 - xy}$.

• $2\tan^{-1} x = \sin^{-1}\dfrac{2x}{1 + x^2}$ for $-1 \le x \le 1$.
• $2\tan^{-1} x = \cos^{-1}\dfrac{1 - x^2}{1 + x^2}$ for $x \ge 0$.
• $2\tan^{-1} x = \tan^{-1}\dfrac{2x}{1 - x^2}$ for $-1 < x < 1$.

Why. Put $x = \tan\theta$. Then $\sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$, $\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$ and $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$. The restrictions on $x$ are exactly what keep $2\theta$ inside the branch of the outer inverse function.

Find $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}$. Here $xy = \frac{1}{6} < 1$, so the rule applies: $\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{5/6}{5/6} = 1$. So the sum is $\tan^{-1} 1 = \frac{\pi}{4}$. A numerical check gives $0.785398$, which is $\frac{\pi}{4}$.

Show that $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$. Using the double-angle form with $x = \frac{1}{3}$ (inside $(-1, 1)$): $\frac{2x}{1 - x^2} = \frac{2/3}{8/9} = \frac{3}{4}$. Both sides are about $0.6435$.

1. Find $\cos^{-1}\!\left(-\frac{1}{2}\right)$ using the rule for negative inputs.
2. Find $\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2} + \tan^{-1} 1$.
3. Show that $2\tan^{-1}\frac{1}{2} = \tan^{-1}\frac{4}{3}$.

Answers

1. $\pi - \cos^{-1}\frac{1}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
2. The first two terms add to $\frac{\pi}{2}$ (complementary pair), and with $\tan^{-1} 1 = \frac{\pi}{4}$ the total is $\frac{3\pi}{4}$.
3. With $x = \frac{1}{2}$: $\frac{2x}{1 - x^2} = \frac{1}{3/4} = \frac{4}{3}$. Both sides are about $0.9273$.

1. Find the value of $\tan^{-1} 1 + \cos^{-1}\!\left(-\frac{1}{2}\right) + \sin^{-1}\!\left(-\frac{1}{2}\right)$.
2. Show that $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{2}{3} = \frac{\pi}{4}$.
3. Show that $2\tan^{-1}\frac{1}{4} = \tan^{-1}\frac{8}{15}$.
4. Find $\sin^{-1}\!\left(\sin\frac{2\pi}{3}\right)$, $\cos^{-1}\!\left(\cos\left(-\frac{\pi}{4}\right)\right)$ and $\tan^{-1}\!\left(\tan\frac{5\pi}{6}\right)$.
5. Find $\cos\!\left(2\sin^{-1}\frac{3}{5}\right)$.
6. Find $\sin\!\left(\tan^{-1}\frac{3}{4} + \cos^{-1}\frac{5}{13}\right)$.
7. For which $x$ is $\tan^{-1} x + \tan^{-1} 2 = \tan^{-1}\frac{x + 2}{1 - 2x}$? Test the formula at $x = 1$.
8. Solve $\tan^{-1} x + \tan^{-1}(2x) = \frac{\pi}{4}$.
9. Show that $\sin^{-1}\!\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1} x$ for $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$. What goes wrong at $x = 0.9$?
10. Show that $\cos(\tan^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$ and $\sin(\tan^{-1} x) = \frac{x}{\sqrt{1 + x^2}}$. Check with $x = \frac{3}{4}$.

Answers

1. The three principal values are $\frac{\pi}{4}$, $\frac{2\pi}{3}$ and $-\frac{\pi}{6}$. Their sum is $\frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.
2. Here $xy = \frac{2}{15} < 1$, so $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{2}{3} = \tan^{-1}\frac{\frac{1}{5} + \frac{2}{3}}{1 - \frac{2}{15}} = \tan^{-1}\frac{13/15}{13/15} = \tan^{-1} 1 = \frac{\pi}{4}$.
3. With $x = \frac{1}{4}$, which is in $(-1, 1)$: $\frac{2x}{1 - x^2} = \frac{1/2}{15/16} = \frac{8}{15}$.
4. Move each angle into its branch. $\sin\frac{2\pi}{3} = \sin\frac{\pi}{3}$, so the first is $\frac{\pi}{3}$. $\cos\left(-\frac{\pi}{4}\right) = \cos\frac{\pi}{4}$ and $\frac{\pi}{4}$ is in $[0, \pi]$, so the second is $\frac{\pi}{4}$. $\tan\frac{5\pi}{6} = \tan\left(-\frac{\pi}{6}\right)$, so the third is $-\frac{\pi}{6}$.
5. Let $\theta = \sin^{-1}\frac{3}{5}$. Then $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \cdot \frac{9}{25} = \frac{7}{25} = 0.28$.
6. Let $A = \tan^{-1}\frac{3}{4}$, so $\sin A = \frac{3}{5}$ and $\cos A = \frac{4}{5}$. Let $B = \cos^{-1}\frac{5}{13}$, so $\cos B = \frac{5}{13}$ and $\sin B = \frac{12}{13}$. Then $\sin(A + B) = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{63}{65} \approx 0.9692$.
7. The formula needs $xy < 1$, that is $2x < 1$, so $x < \frac{1}{2}$. At $x = 1$ the left side is $1.8925$ and the right side is $-1.249$. They differ by $\pi$, so the formula fails there.
8. Using the sum rule: $\frac{3x}{1 - 2x^2} = 1$, so $2x^2 + 3x - 1 = 0$ and $x = \frac{-3 \pm \sqrt{17}}{4}$, that is $x \approx 0.2808$ or $x \approx -1.7808$. The rule needs $2x^2 < 1$. This holds for the first root, and putting it back gives $\frac{\pi}{4}$. For the second root both angles are negative and $2x^2 > 1$; the left side is actually $-2.3562$, which is $-\frac{3\pi}{4}$. So the only solution is $x = \frac{-3 + \sqrt{17}}{4}$.
9. Put $x = \sin\theta$ with $\theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$. Then $2x\sqrt{1 - x^2} = 2\sin\theta\cos\theta = \sin 2\theta$, and $2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is in the branch, so $\sin^{-1}(\sin 2\theta) = 2\theta$. At $x = 0.5$ both sides are $1.0472$. At $x = 0.9$ the angle $2\theta$ is bigger than $\frac{\pi}{2}$, so the left side is $0.9021$ while the right side is $2.2395$.
10. Let $\theta = \tan^{-1} x$, which lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $\cos\theta > 0$. In a right triangle with opposite $x$ and adjacent $1$, the hypotenuse is $\sqrt{1 + x^2}$. For $x = \frac{3}{4}$: $\cos\theta = \frac{1}{\sqrt{1 + 9/16}} = \frac{4}{5} = 0.8$.

Show that $\tan^{-1}\frac{1}{4} + \tan^{-1}\frac{3}{5} = \frac{\pi}{4}$. Here $xy = \frac{3}{20} < 1$, so the sum rule works: $\frac{\frac{1}{4} + \frac{3}{5}}{1 - \frac{3}{20}} = \frac{17/20}{17/20} = 1$, and $\tan^{-1} 1 = \frac{\pi}{4}$.

Find $\sin\!\left(2\cos^{-1}\!\left(-\frac{3}{5}\right)\right)$. Let $\theta = \cos^{-1}\!\left(-\frac{3}{5}\right)$. The cosine is negative, so $\theta$ is in $\left(\frac{\pi}{2}, \pi\right)$, where sine is positive. From $\cos\theta = -\frac{3}{5}$ we get $\sin\theta = \frac{4}{5}$. Then $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{4}{5} \cdot \left(-\frac{3}{5}\right) = -\frac{24}{25} = -0.96$.

Solve $\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}$.

By the sum rule, $\frac{5x}{1 - 6x^2} = 1$, so $6x^2 + 5x - 1 = 0$. Factorising, $(6x - 1)(x + 1) = 0$, so $x = \frac{1}{6}$ or $x = -1$. Check each answer, because the sum rule needs $6x^2 < 1$. For $x = \frac{1}{6}$: $6x^2 = \frac{1}{6} < 1$, and indeed $\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2} = \frac{\pi}{4}$. For $x = -1$: $6x^2 = 6 > 1$, and the left side is $\tan^{-1}(-2) + \tan^{-1}(-3) = -2.3562$, which is $-\frac{3\pi}{4}$, not $\frac{\pi}{4}$. So the only solution is $x = \frac{1}{6}$.

A picture $2$ m tall hangs on a wall so that its bottom edge is $1$ m above your eye level. You stand $3$ m from the wall. At what angle does the picture appear to you?

The angle to the top edge is $\tan^{-1}\frac{3}{3} = \tan^{-1} 1 = 45°$. The angle to the bottom edge is $\tan^{-1}\frac{1}{3} \approx 18.43°$. The picture fills the difference. By the difference rule, $\tan^{-1} 1 - \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{1 - 1/3}{1 + 1/3} = \tan^{-1}\frac{1}{2} \approx 26.57°$.

Is $\sin^{-1}(\sin 2) = 2$? No. The angle $2$ radians is about $114.6°$, which is outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\sin 2 = \sin(\pi - 2)$ and $\pi - 2 \approx 1.1416$ is in the branch, $\sin^{-1}(\sin 2) = \pi - 2 \approx 1.1416$.

1. Find $\cos^{-1}\!\left(\cos\frac{5\pi}{3}\right)$.
2. Show that $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$.
3. Show that $\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$.
4. Find the value of $\sin^{-1} x + \cos^{-1} x + \tan^{-1} x$ when $x = 1$.
5. A student writes $\cos^{-1}\!\left(-\frac{1}{2}\right) = \frac{4\pi}{3}$ because $\cos\frac{4\pi}{3} = -\frac{1}{2}$. What is wrong, and what is the correct answer?
6. Find the domain and range of $f(x) = \sin^{-1}(2x - 1)$.
7. A ramp rises $3$ m over a flat distance of $4$ m. Find the angle of the ramp with the ground, to one decimal place, and the length of the ramp.
8. Show that $\cos^{-1} x = 2\sin^{-1}\sqrt{\frac{1 - x}{2}}$ for $0 \le x \le 1$.

Answers

1. $\frac{5\pi}{3}$ is outside $[0, \pi]$. Since $\cos\frac{5\pi}{3} = \frac{1}{2}$, the answer is $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$.
2. Let $\theta = \sin^{-1}\frac{3}{5}$, so $\tan\theta = \frac{3}{4}$. Then $\tan 2\theta = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{3/2}{7/16} = \frac{24}{7}$. Since $\theta < \frac{\pi}{4}$ is not needed here (we have $\tan\theta = \frac{3}{4} < 1$, so $\theta < \frac{\pi}{4}$), $2\theta < \frac{\pi}{2}$ is in the branch of $\tan^{-1}$.
3. Add in pairs. $\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} = \tan^{-1}\frac{8/15}{14/15} = \tan^{-1}\frac{4}{7}$. $\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{15/56}{55/56} = \tan^{-1}\frac{3}{11}$. Then $\tan^{-1}\frac{4}{7} + \tan^{-1}\frac{3}{11} = \tan^{-1}\frac{65/77}{65/77} = \tan^{-1} 1 = \frac{\pi}{4}$. Every step has $xy < 1$.
4. $\sin^{-1} 1 + \cos^{-1} 1 + \tan^{-1} 1 = \frac{\pi}{2} + 0 + \frac{\pi}{4} = \frac{3\pi}{4}$. (The first two always add to $\frac{\pi}{2}$.)
5. The value $\frac{4\pi}{3}$ is outside the principal branch $[0, \pi]$ of $\cos^{-1}$. The correct answer is $\frac{2\pi}{3}$.
6. We need $-1 \le 2x - 1 \le 1$, so $0 \le x \le 1$. This is the domain. At $x = 0$, $f = -1.5708$ (that is $-\frac{\pi}{2}$), and at $x = 1$, $f = 1.5708$ (that is $\frac{\pi}{2}$), and $f$ rises in between. The range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
7. The angle is $\tan^{-1}\frac{3}{4} \approx 36.9°$. The length is $\sqrt{3^2 + 4^2} = 5$ m.
8. Let $\alpha = \cos^{-1} x$, which is in $\left[0, \frac{\pi}{2}\right]$ because $x \ge 0$. Then $\sin^2\frac{\alpha}{2} = \frac{1 - \cos\alpha}{2} = \frac{1 - x}{2}$. Since $\frac{\alpha}{2} \in \left[0, \frac{\pi}{4}\right]$ is in the branch of $\sin^{-1}$, $\frac{\alpha}{2} = \sin^{-1}\sqrt{\frac{1 - x}{2}}$. Multiply by $2$.