Trigonometry

Inverse Trig Functions

You know that sin(30 degrees) = 0.5. But what if someone gives you the answer (0.5) and asks for the angle? That’s what inverse trig functions do — they go backwards from a ratio to an angle. The catch? Trig functions repeat forever, so we need to be very careful about which angle we pick.


1. Why Sin(x) Doesn’t Have an Inverse (At First)

Let’s apply the horizontal line test to sin(x). Move the line up and down and count how many times it crosses the sine curve.

Horizontal line y = ?0.5
-11
-10-8-6-4-2246810-22y = sin(x)y = test line
Connection

The horizontal line hits the sine curve in infinitely many places! For example, sin(x) = 0.5 has solutions at x = pi/6, 5pi/6, 13pi/6, and so on forever. That means sin(x) is NOT one-to-one over all real numbers. To create an inverse, we must restrict the domain to an interval where sin is one-to-one.


2. The Restricted Sine and Arcsin

We restrict sine to the interval [-pi/2, pi/2], where it goes from -1 up to 1 without repeating. On this interval it passes the horizontal line test, so the inverse exists. This inverse is called arcsin (or sin-inverse).

Input value0.5
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arcsin(0.5)=angle in [π0,π0]\arcsin(0.5) = \text{angle in } [-\tfrac{\pi}0, \tfrac{\pi}0]
-2-1.5-1-0.50.511.52-2-1.5-1-0.50.511.52y = arcsin(x)y = xinput value
Try This

Move the input slider. Notice how arcsin only accepts inputs between -1 and 1 (the range of sine), and only outputs angles between -pi/2 and pi/2 (about -1.57 to 1.57). The curve exists only in this rectangle — that’s the restricted domain and range at work.


3. Arccos — The Cosine Inverse

For cosine, we restrict the domain to [0, pi], where cos goes from 1 down to -1 monotonically. The inverse is arccos.

Input to arccos0.5
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arccos(0.5)=angle in [0,π]\arccos(0.5) = \text{angle in } [0, \pi]
-2-1.5-1-0.50.511.52-0.50.511.522.533.54y = arccos(x)input value
Connection

Arccos has the same domain as arcsin (inputs between -1 and 1), but its range is different: it outputs angles from 0 to pi (about 0 to 3.14). Notice that arccos is a decreasing function — bigger inputs give smaller angles. That’s because cosine decreases on [0, pi].


4. Arctan — The Tangent Inverse

Tangent is restricted to (-pi/2, pi/2), where it sweeps from negative infinity to positive infinity. The inverse, arctan, accepts any real number and outputs an angle in that range.

Input to arctan1
-1010
arctan(1)=angle in (π0,π0)\arctan(1) = \text{angle in } (-\tfrac{\pi}0, \tfrac{\pi}0)
-12-10-8-6-4-224681012-22y = arctan(x)input value
Try This

Arctan is defined for ALL real numbers — you can slide the input anywhere. But the output never exceeds pi/2 or drops below -pi/2. Those horizontal boundaries are called asymptotes of the arctan curve. No matter how large the input, the angle approaches but never reaches 90 degrees.


5. Comparing All Three Inverse Trig Functions

Let’s see arcsin, arccos, and arctan on the same graph to compare their domains, ranges, and shapes.

Shared input (for arcsin & arccos)0.5
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-2-1.5-1-0.50.511.52-2-1.5-1-0.50.511.522.533.54arcsin(x)arccos(x)arctan(x)input line
Connection

At the same input value, the three functions give different angles because they’re answering different questions. Notice that arcsin(x) + arccos(x) = pi/2 for every valid input. That’s because sin(theta) = cos(pi/2 - theta), so if theta answers one, pi/2 - theta answers the other.


6. From Ratio Back to Angle

Here’s the practical use case. Given a sine value, arcsin tells you the angle. Let’s trace this visually: pick a y-value on the restricted sine curve, and see the corresponding angle.

Sine ratio0.5
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sin(θ)=0.5    θ=arcsin(0.5)\sin(\theta) = 0.5 \implies \theta = \arcsin(0.5)
-2-1.5-1-0.50.511.52-1.5-1-0.50.511.5y = sin(x) restrictedy = ratio
Challenge

Challenge: Using the slider, find the angle whose sine is 0.866. You should get approximately pi/3 (about 1.047). Now find the angle whose sine is -0.707. That should be approximately -pi/4 (about -0.785). In both cases, arcsin gives the unique answer in [-pi/2, pi/2]. If you needed an answer in a different quadrant, you’d use reference angles and the unit circle.

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