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Inverse Trigonometry

Inverse Trigonometry
Inverse Trig

The study of the inverses of the six trig functions.

Technical note: Since none of the six trig functions sine, cosine, tangent, cosecant, secant, and cotangent are one-to-one, their inverses are not functions. Each trig function can have its domain restricted, however, in order to make its inverse a function. Some mathematicians write these restricted trig functions and their inverses with an initial capital letter (e.g. Cos or Cos-1). However, most mathematicians do not follow this practice. This website does not distinguish between capitalized and uncapitalized trig functions.

 

 

See also

Inverse trig functions

Key Formula

If y=sin⁡(x), then x=sin⁡−1(y)=arcsin⁡(y)\text{If } y = \sin(x), \text{ then } x = \sin^{-1}(y) = \arcsin(y)
Where:
  • yy = A ratio value (between −1 and 1 for sine and cosine)
  • xx = The angle (in radians or degrees) that produces that ratio

Worked Example

Problem: Find the angle θ such that sin(θ) = 0.5, using inverse trigonometry.
Step 1: Identify that you need to reverse the sine function, so apply arcsin to both sides.
θ=sin⁡−1(0.5)\theta = \sin^{-1}(0.5)
Step 2: Recall the restricted range of arcsin. The output of arcsin is always in the interval [−π2, π2][-\frac{\pi}{2},\, \frac{\pi}{2}], which is [−90°,90°][-90°, 90°].
Step 3: Evaluate. Since sin⁡(30°)=0.5\sin(30°) = 0.5, and 30°30° falls within the restricted range, the answer is:
θ=sin⁡−1(0.5)=30°=π6 radians\theta = \sin^{-1}(0.5) = 30° = \frac{\pi}{6} \text{ radians}
Answer: θ = 30° (or π/6 radians)

Another Example

Problem: Find the angle whose tangent is −1, i.e., evaluate tan⁡−1(−1)\tan^{-1}(-1).
Step 1: Apply the inverse tangent function.
θ=tan⁡−1(−1)\theta = \tan^{-1}(-1)
Step 2: Recall the restricted range of arctan: (−π2, π2)(-\frac{\pi}{2},\, \frac{\pi}{2}), or (−90°,90°)(-90°, 90°).
Step 3: Since tan⁡(−45°)=−1\tan(-45°) = -1 and −45°-45° is within the allowed range:
θ=tan⁡−1(−1)=−45°=−π4 radians\theta = \tan^{-1}(-1) = -45° = -\frac{\pi}{4} \text{ radians}
Answer: θ = −45° (or −π/4 radians)

Frequently Asked Questions

What is the difference between arcsin and sin⁻¹?
They mean exactly the same thing. The notation arcsin⁡(x)\arcsin(x) and sin⁡−1(x)\sin^{-1}(x) both refer to the inverse sine function. Be careful not to confuse sin⁡−1(x)\sin^{-1}(x) with 1sin⁡(x)\frac{1}{\sin(x)}, which is the cosecant function csc⁡(x)\csc(x).
Why do inverse trig functions have restricted ranges?
Trig functions are periodic, meaning many different angles produce the same ratio. For example, both 30° and 150° have a sine of 0.5. To make the inverse a proper function (one output per input), mathematicians restrict the range so each input gives exactly one angle. For arcsin, the range is [−90°,90°][-90°, 90°]; for arccos, it is [0°,180°][0°, 180°]; and for arctan, it is (−90°,90°)(-90°, 90°).

Trig Functions vs. Inverse Trig Functions

Trig functions take an angle as input and output a ratio (e.g., sin⁡(30°)=0.5\sin(30°) = 0.5). Inverse trig functions do the opposite: they take a ratio as input and output an angle (e.g., sin⁡−1(0.5)=30°\sin^{-1}(0.5) = 30°). They undo each other, but only within the restricted domains and ranges.

Why It Matters

Inverse trig functions are essential whenever you need to find an unknown angle. In physics, you use tan⁡−1\tan^{-1} to find the direction of a resultant vector from its components. In geometry and engineering, arcsin and arccos help you solve triangles when you know side lengths but not the angles. Calculators and programming languages all include these functions as built-in tools.

Common Mistakes

Mistake: Confusing sin⁡−1(x)\sin^{-1}(x) with 1sin⁡(x)\frac{1}{\sin(x)}.
Correction: The notation sin⁡−1(x)\sin^{-1}(x) means the inverse sine (arcsin), not the reciprocal. The reciprocal of sine is cosecant: csc⁡(x)=1sin⁡(x)\csc(x) = \frac{1}{\sin(x)}. The −1-1 here is not an exponent — it signals an inverse function.
Mistake: Forgetting about restricted ranges and giving extra solutions.
Correction: Each inverse trig function returns exactly one value from its principal range. For instance, cos⁡−1(−0.5)=120°\cos^{-1}(-0.5) = 120°, not 240°240°. If a problem asks for all solutions, you must use the principal value along with the periodicity of the original trig function to find the others.

Related Terms