The Principal Values of the Inverse Functions

Many angles, one answer

The equation $\sin \theta =0.5$sin θ = 0.5 has many solutions. Between 0° and 360° there are two, 30° and 150°. A full turn brings each point on the unit circle back to the same place, so 30° + 360° = 390° and 150° + 360° = 510° are solutions too, and so are 30° − 360° = −330° and 150° − 360° = −210°. There are infinitely many.

A calculator key can only show one angle. So each inverse function, ${\sin }^{-1}$sin⁻¹, ${\cos }^{-1}$cos⁻¹ and ${\tan }^{-1}$tan⁻¹, is defined to give one agreed angle. That angle is called the principal value. ${\sin }^{-1}\left(0.5\right)=30^{\circ }$sin⁻¹(0.5) = 30°, and 30° is the principal value of ${\sin }^{-1}\left(0.5\right)$sin⁻¹(0.5).

The level line at height 0.5 cuts the circle at 30° and 150°, and cuts the wave twice in every turn. Turn n to move both solutions a whole turn along, to 390° and 510° or to −330° and −210°. Every dot on the wave has a sine of 0.5.

The window for ${\sin }^{-1}$sin⁻¹

To pick one angle for each value, choose a stretch of angles, a window, in which the sine takes every value from −1 to 1 exactly once. Then every value has exactly one angle in the window, and that angle is the one ${\sin }^{-1}$sin⁻¹ gives.

The window for ${\sin }^{-1}$sin⁻¹ is −90° to 90°. Across it the graph of $y=\sin \theta$y = sin θ rises steadily from −1 at −90°, through 0 at 0°, to 1 at 90°, so it reaches each value once. The window also holds every acute angle, so ${\sin }^{-1}$sin⁻¹ gives the same answer here as it did for an angle in a right triangle.

A positive value gives an angle between 0° and 90°, and a negative value gives an angle between −90° and 0°: ${\sin }^{-1}\left(-0.5\right)=-30^{\circ }$sin⁻¹(−0.5) = −30°.

The graph of $y=\sin \theta$y = sin θ from −90° to 90°, one square across for every 90°. It rises from −1 to 1 and meets every height between them exactly once.

The window for ${\cos }^{-1}$cos⁻¹

The same window does not work for the cosine. Between −90° and 90° the cosine rises from 0 to 1 and falls back to 0, so cos 60° and cos (−60°) are both ½, and no angle in that window has a negative cosine.

The window for ${\cos }^{-1}$cos⁻¹ is 0° to 180°. Across it the graph of $y=\cos \theta$y = cos θ falls steadily from 1 at 0°, through 0 at 90°, to −1 at 180°, so it reaches each value from −1 to 1 once. It also holds every acute angle.

So ${\cos }^{-1}$cos⁻¹ never gives a negative angle. A positive value gives an angle between 0° and 90°, and a negative value an angle between 90° and 180°: ${\cos }^{-1}\left(-0.5\right)=120^{\circ }$cos⁻¹(−0.5) = 120°.

The graph of $y=\cos \theta$y = cos θ from 0° to 180°, one square across for every 90°. It falls from 1 to −1 and meets every height between them exactly once.

The window for ${\tan }^{-1}$tan⁻¹

The tangent takes every value, from very large negative numbers to very large positive ones, on the single branch of its graph between −90° and 90°. So the window for ${\tan }^{-1}$tan⁻¹ is $-90^{\circ }<\theta <90^{\circ }$−90° < θ < 90°. The ends are left out because tan 90° and tan (−90°) do not exist.

Like ${\sin }^{-1}$sin⁻¹, ${\tan }^{-1}$tan⁻¹ gives a negative angle for a negative value: ${\tan }^{-1}\left(-1\right)=-45^{\circ }$tan⁻¹(−1) = −45°.

In short: ${\sin }^{-1}$sin⁻¹ gives an angle from −90° to 90°, ${\cos }^{-1}$cos⁻¹ an angle from 0° to 180°, and ${\tan }^{-1}$tan⁻¹ an angle between −90° and 90°. The inverse functions are also written arcsin, arccos and arctan.

The principal value, then the symmetry

To solve an equation in a range, take the principal value from the inverse, then use the symmetry of the graph for the other angles, then add or subtract 360° until every angle in the range is found.

Solve $\sin \theta =0.5$sin θ = 0.5 for $0^{\circ }\le \theta \le 360^{\circ }$0° ≤ θ ≤ 360°. The principal value is ${\sin }^{-1}\left(0.5\right)=30^{\circ }$sin⁻¹(0.5) = 30°. The sine graph is symmetrical about 90°, so the second angle is 180° − 30° = 150°. The solutions are 30° and 150°.

Solve $\cos \theta =-0.5$cos θ = −0.5 for $0^{\circ }\le \theta \le 360^{\circ }$0° ≤ θ ≤ 360°. The principal value is ${\cos }^{-1}\left(-0.5\right)=120^{\circ }$cos⁻¹(−0.5) = 120°. The cosine graph is symmetrical about 180°, so the second angle is 360° − 120° = 240°. Check: cos 240° = −0.5.

Solve $\sin \theta =-0.3$sin θ = −0.3 for $0^{\circ }\le \theta \le 360^{\circ }$0° ≤ θ ≤ 360°. The principal value is ${\sin }^{-1}\left(-0.3\right)=-17.46^{\circ }$sin⁻¹(−0.3) = −17.46°, to 2 decimal places, which is outside the range. Adding 360° gives 342.54°. The second angle is 180° − (−17.46°) = 197.46°. The solutions are 197.46° and 342.54°.

Solve $\tan \theta =-1$tan θ = −1 for $0^{\circ }\le \theta \le 360^{\circ }$0° ≤ θ ≤ 360°. The principal value is ${\tan }^{-1}\left(-1\right)=-45^{\circ }$tan⁻¹(−1) = −45°, outside the range. The tangent repeats every 180°, so the solutions are −45° + 180° = 135° and −45° + 360° = 315°.

The inverse undoes the ratio only inside its window

$\sin \left({\sin }^{-1}0.5\right)=\sin 30^{\circ }=0.5$sin (sin⁻¹ 0.5) = sin 30° = 0.5, so taking the sine of the inverse sine gives the number back. The other order does not always work. sin 150° = 0.5, so ${\sin }^{-1}\left(\sin 150^{\circ }\right)={\sin }^{-1}\left(0.5\right)=30^{\circ }$sin⁻¹(sin 150°) = sin⁻¹(0.5) = 30°, not 150°.

The inverse returns the angle in its window, whatever angle the sine came from. That is why the second angle, and every angle a whole turn away, has to come from the symmetry and not from the calculator.

The usual mistakes

Giving an angle outside the window as the value of the inverse. 150° has a sine of 0.5, but ${\sin }^{-1}\left(0.5\right)$sin⁻¹(0.5) is 30°, because 150° is not between −90° and 90°.

Dropping the sign. ${\sin }^{-1}\left(-0.5\right)$sin⁻¹(−0.5) is −30° and ${\tan }^{-1}\left(-1\right)$tan⁻¹(−1) is −45°. For ${\cos }^{-1}a$cos⁻¹ a negative value gives an obtuse angle: ${\cos }^{-1}\left(-0.5\right)=120^{\circ }$cos⁻¹(−0.5) = 120°, not −120° and not 60°.

Stopping at the principal value. In a range such as 0° to 360°, the equation usually has a second solution, and the principal value may itself be outside the range.

When the angle is 12x

In the application below, the height of a Ferris wheel rider x minutes after boarding is h = 20 − 18 cos (12x)°. Setting h = 29 gives cos (12x)° = −½, an equation in the angle 12x, not in x.

Find every value of the angle 12x first, and only then divide by 12. The first hour is $0\le x\le 60$0 ≤ x ≤ 60, so the angle 12x runs from 12 × 0 = 0° to 12 × 60 = 720°, which is two full turns. Each turn holds two angles with a cosine of −½, the principal value and its pair, so there are four values of 12x, and four times.