Inverse Trigonometric Functions

Undoing sine

Sine turns an angle into a height on the unit circle. Undoing it means going from the height back to the angle: which angle has a sine of $\tfrac{1}{2}$?

There are too many answers. $30°$ has that sine, and so do $150°$, $390°$, $510°$ and $-210°$. The graph shows why: a horizontal line at height $\tfrac{1}{2}$ crosses the sine wave over and over.

• y = sin x
• sin x on −π/2 ≤ x ≤ π/2

A function whose outputs repeat fails the horizontal line test, so it has no inverse function as it stands.

Why the domain has to be restricted

The repair is the one used for $x^2$: keep only part of the graph. The part has to do two jobs.

Each output only once. The piece must always rise or always fall. A piece that turned around would repeat outputs, and the inverse would be two-valued.

Every output at least once. The piece should still reach every sine value from $-1$ to $1$, so the inverse accepts every input it could be given.

The piece from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ does both. It rises the whole way, from $-1$ to $1$, and passes through $0$ at the origin. So restricted there, sine has an inverse function, called arcsine or $\sin^{-1}$:

$$\arcsin x = \theta \quad\text{means}\quad \sin\theta = x \;\text{ and }\; -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$$

Its graph is the restricted piece reflected across $y = x$, as every inverse is.

• y = sin x, restricted
• y = arcsin x
• y = x

The restriction is a choice, but a forced one. Any interval where sine rises through every value once would work. This one is chosen because it contains the small positive angles, where $\arcsin$ agrees with right-triangle trigonometry.

Cosine and tangent

Cosine is not one-to-one on $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$, since $\cos(-x) = \cos x$. It needs a different piece: from $0$ to $\pi$ it falls steadily from $1$ to $-1$. Tangent rises on $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ through every real number.

Inverse	Accepts	Returns an angle in
$\arcsin x$	$-1 \le x \le 1$	$-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$
$\arccos x$	$-1 \le x \le 1$	$0 \le \theta \le \pi$
$\arctan x$	every real number	$-\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}$

Solving trigonometric equations

An inverse function returns one angle. An equation usually has more solutions, and the unit circle supplies them.

Solve $\sin x = 0.3$ for $0 \le x < 2\pi$.

$$x = \arcsin 0.3 \approx 0.3047$$

Sine is a height, and the point at the same height on the other side of the circle has angle $\pi - x$:

$$x = \pi - 0.3047 \approx 2.8369$$

Both are solutions, and adding any multiple of $2\pi$ to either gives the rest. For cosine the partner comes from the other reflection: if $\cos x = c$, then $\cos(2\pi - x) = c$ too.

In a model

A harbor’s depth is $d(t) = 8 + 4\cos\left(\tfrac{\pi t}{6}\right)$ feet, $t$ hours after midnight, from graphing sine and cosine. When is the water $5$ feet deep?

$$8 + 4\cos\left(\tfrac{\pi t}{6}\right) = 5 \quad\Rightarrow\quad \cos\left(\tfrac{\pi t}{6}\right) = -0.75$$ $$\tfrac{\pi t}{6} = \arccos(-0.75) \approx 2.4189 \quad\Rightarrow\quad t \approx 4.62$$

The cosine repeats its values symmetrically within each $12$-hour cycle, so the other time is $12 - 4.62 = 7.38$. The water is $5$ feet deep at about $4{:}37$ a.m., falling, and at about $7{:}23$ a.m., rising again.

Interpreting is part of the answer. The inverse gave a number; the model gave it a time of day, and the symmetry of the tide found the second one.

Worked examples

Common mistakes

Practice problems