Learn arcsin, arccos, and arctan — the functions that reverse sine, cosine, and tangent to find angles from ratios
Trigonometric functions like sine, cosine, and tangent take an angle as input and produce a ratio (a number) as output. But what if you know the ratio and need to find the angle? That is the job of the inverse trig functions.
For example, if you know that sin(θ) = 0.5, what is θ? You need a function that takes 0.5 as input and returns the angle. That function is arcsin (also written sin−1). So arcsin(0.5) = 30° (or π/6 radians).
Inverse trig functions are essential for solving trigonometric equations, finding angles in right-triangle problems, converting between rectangular and polar coordinates, and working with integrals in calculus.
There are two common notations for inverse trig functions, and you will encounter both:
Both notations mean exactly the same thing. However, it is critical to understand that sin−1(x) does NOT mean 1/sin(x). The notation sin−1(x) is the inverse function (finding the angle), while 1/sin(x) is the reciprocal (which is csc(x)). This is the single most common source of confusion with inverse trig notation.
sin−1(x) = arcsin(x) = "the angle whose sine is x." It is NOT 1/sin(x). The reciprocal of sine is cosecant (csc), which is a completely different function.
Here is the fundamental problem with defining inverse trig functions: trig functions are not one-to-one. For example, sin(30°) = 0.5, but sin(150°) = 0.5 too. In fact, there are infinitely many angles whose sine is 0.5. So when we ask "what is arcsin(0.5)?", which answer do we give?
The solution is to restrict the domain of each trig function to an interval where it is one-to-one (passes the horizontal line test). The output of the inverse function is then called the principal value.
| Function | Input (Domain) | Output (Range / Principal Values) |
|---|---|---|
| arcsin(x) | [-1, 1] | [-π/2, π/2] or [-90°, 90°] |
| arccos(x) | [-1, 1] | [0, π] or [0°, 180°] |
| arctan(x) | (-∞, +∞) | (-π/2, π/2) or (-90°, 90°) |
These restrictions are chosen carefully:
The arcsin graph is defined only for x in [-1, 1] and produces y values from −π/2 to π/2. It is an increasing function that passes through the origin.
The arccos graph is also defined for x in [-1, 1] but produces y values from 0 to π. It is a decreasing function. At x = 1, arccos(1) = 0; at x = −1, arccos(−1) = π.
The arctan graph accepts all real numbers as input and has two horizontal asymptotes at y = π/2 and y = −π/2 (which it approaches but never reaches). It is an increasing function that passes through the origin and is widely used in calculus and physics.
Each inverse trig function is the reflection of its corresponding trig function over the line y = x, restricted to a specific interval. The restricted ranges ensure that each input gives exactly one output.
| Input x | arcsin(x) | arccos(x) | arctan(x) |
|---|---|---|---|
| -1 | -π/2 | π | -π/4 |
| -√3/2 | -π/3 | 5π/6 | — |
| -√2/2 | -π/4 | 3π/4 | — |
| -1/2 | -π/6 | 2π/3 | — |
| 0 | 0 | π/2 | 0 |
| 1/2 | π/6 | π/3 | — |
| √2/2 | π/4 | π/4 | — |
| √3/2 | π/3 | π/6 | — |
| 1 | π/2 | 0 | π/4 |
Composing a trig function with its inverse (or vice versa) requires careful attention to the restricted domains:
These work because applying a function and then its inverse brings you back to where you started.
If x is outside the restricted range, the inverse function will return the principal value equivalent, not the original x. For example, arcsin(sin(5π/6)) = arcsin(1/2) = π/6, NOT 5π/6, because π/6 is the principal value in [-π/2, π/2].
Expressions like cos(arctan(x)) or sin(arccos(x)) involve composing a trig function with a different inverse trig function. These can be evaluated using right triangles:
To evaluate a mixed composition like cos(arctan(x)), draw a right triangle. If arctan(x) = θ, then tan(θ) = x = x/1, so the opposite side is x and the adjacent side is 1. The hypotenuse is √(x² + 1). Then cos(θ) = adjacent/hypotenuse = 1/√(x² + 1).
Every scientific calculator has sin−1, cos−1, and tan−1 buttons (often accessed by pressing a SHIFT or 2ND key before the sin, cos, or tan button). Here are some important tips:
Inverse trig functions appear in many contexts beyond basic triangle problems:
Step 1: We need the angle θ in [-π/2, π/2] such that sin(θ) = 1/2.
Step 2: From the unit circle, sin(30°) = sin(π/6) = 1/2.
Step 3: Since π/6 is within the range [-π/2, π/2], it is the principal value.
Answer: arcsin(1/2) = π/6 (or 30°)
Step 1: Let θ = arctan(3/4). This means tan(θ) = 3/4, where θ is in (-π/2, π/2).
Step 2: Draw a right triangle where the opposite side = 3 and the adjacent side = 4.
Step 3: Find the hypotenuse: h = √(3² + 4²) = √(9 + 16) = √25 = 5.
Step 4: cos(θ) = adjacent/hypotenuse = 4/5.
Answer: cos(arctan(3/4)) = 4/5 (or 0.8)
Step 1: Find the principal value: x₁ = arcsin(0.6) ≈ 0.6435 radians (about 36.87°).
Step 2: Since sine is also positive in Quadrant II, the second solution is: x₂ = π − 0.6435 ≈ 2.4981 radians (about 143.13°).
Step 3: Verify: sin(0.6435) ≈ 0.6 and sin(2.4981) ≈ 0.6. Both check out.
Answer: x ≈ 0.6435 and x ≈ 2.4981 radians
Find the exact value of arccos(−√2/2).
We need θ in [0, π] where cos(θ) = −√2/2. The reference angle for √2/2 is π/4. Cosine is negative in Quadrant II, so θ = π − π/4 = 3π/4 (or 135°).
Use a right triangle to find the exact value of sin(arccos(5/13)).
Let θ = arccos(5/13), so cos(θ) = 5/13. The adjacent side is 5 and the hypotenuse is 13. By the Pythagorean theorem, the opposite side = √(13² − 5²) = √(169 − 25) = √144 = 12. Therefore sin(θ) = 12/13. Answer: 12/13.
Find all solutions to cos(x) = −0.3 in the interval [0, 2π).
Principal value: x₁ = arccos(−0.3) ≈ 1.8755 radians (about 107.46°). Cosine is also negative in Quadrant III. The second solution is: x₂ = 2π − 1.8755 ≈ 4.4077 radians (about 252.54°). Answer: x ≈ 1.8755 and x ≈ 4.4077 radians.
Inverse trig functions are the key to going from ratios back to angles. Always remember: they return only one value (the principal value). To find all solutions to a trig equation, you must use the properties of the trig function (symmetry, periodicity) to find the additional solutions.