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Inverse Sine

Inverse Sine
sin-1
Sin-1
arcsin
Arcsin

The inverse function of sine.

Basic idea: To find sin-1(½), we ask "what angle has sine equal to ½?" The answer is 30°. As a result we say sin-1(½) = 30°. In radians this is sin-1(½) = π/6.

More: There are actually many angles that have sine equal to ½. We are really asking "what is the simplest, most basic angle that has sine equal to ½?" As before, the answer is 30°. Thus sin-1(½) = 30° or sin-1(½) = π/6.

Details: What is sin-1(–½)? Do we choose 210°, –30°, 330° , or some other angle? The answer is –30°. With inverse sine, we select the angle on the right half of the unit circle having measure as close to zero as possible. Thus sin-1(–½) = –30° or sin–1(–½) = –π/6.

In other words, the range of sin-1 is restricted to [–90°, 90°] or The interval [-π/2, π/2].

Note: arcsin refers to "arc sine", or the radian measure of the arc on a circle corresponding to a given value of sine.

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. Sin or Sin-1). However, most mathematicians do not follow this practice. This website does not distinguish between capitalized and uncapitalized trig functions.

 

Unit circle showing range of sin⁻¹: from -π/2 (bottom) to π/2 (top), with 0 on the right.

 

 

See also

Inverse trigonometry, inverse trig functions, interval notation

Key Formula

y=sin⁡−1(x)⟺sin⁡(y)=xy = \sin^{-1}(x) \quad \Longleftrightarrow \quad \sin(y) = x
Where:
  • xx = The input value (a ratio), which must satisfy −1 ≤ x ≤ 1
  • yy = The output angle, restricted to −π/2 ≤ y ≤ π/2 (or −90° ≤ y ≤ 90°)

Worked Example

Problem: Find sin⁻¹(√3/2) in both degrees and radians.
Step 1: Set up the equation. You need to find the angle y such that sin(y) equals √3/2.
sin⁡(y)=32\sin(y) = \frac{\sqrt{3}}{2}
Step 2: Recall which standard angle has a sine of √3/2. From the unit circle, sin(60°) = √3/2.
sin⁡(60°)=sin⁡ ⁣(π3)=32\sin(60°) = \sin\!\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
Step 3: Check that 60° falls within the required range of [−90°, 90°]. It does, so this is the correct answer.
−90°≤60°≤90°✓-90° \leq 60° \leq 90° \quad \checkmark
Step 4: State the result in both degrees and radians.
sin⁡−1 ⁣(32)=60°=π3\sin^{-1}\!\left(\frac{\sqrt{3}}{2}\right) = 60° = \frac{\pi}{3}
Answer: sin⁻¹(√3/2) = 60° = π/3

Another Example

This example uses a negative input, showing that inverse sine returns a negative angle for negative inputs. It also illustrates why you must reject angles like 225° or 315° that lie outside the restricted range.

Problem: Find sin⁻¹(−√2/2) in both degrees and radians.
Step 1: Set up the equation. Find the angle y such that sin(y) = −√2/2.
sin⁡(y)=−22\sin(y) = -\frac{\sqrt{2}}{2}
Step 2: Recall that sin(45°) = √2/2. Since you need a negative sine value, the angle must be negative (in the fourth quadrant of the unit circle).
sin⁡(−45°)=−22\sin(-45°) = -\frac{\sqrt{2}}{2}
Step 3: Check the range. The angle −45° lies within [−90°, 90°], so it is valid. Note that 225° and 315° also have sine equal to −√2/2, but they fall outside the restricted range and are not accepted.
−90°≤−45°≤90°✓-90° \leq -45° \leq 90° \quad \checkmark
Step 4: Write the final answer in both units.
sin⁡−1 ⁣(−22)=−45°=−π4\sin^{-1}\!\left(-\frac{\sqrt{2}}{2}\right) = -45° = -\frac{\pi}{4}
Answer: sin⁻¹(−√2/2) = −45° = −π/4

Frequently Asked Questions

What is the difference between arcsin and sin⁻¹?
There is no difference—arcsin(x) and sin⁻¹(x) are two notations for the same function. The notation "arcsin" comes from the idea of finding the arc length on a unit circle. The notation sin⁻¹ uses the standard superscript convention for inverse functions. Be careful: sin⁻¹(x) does not mean 1/sin(x); that would be csc(x).
Why is the range of inverse sine restricted to [−90°, 90°]?
Sine is not one-to-one over all real numbers, so infinitely many angles share the same sine value. To make the inverse a proper function (one output for each input), mathematicians restrict the output to [−90°, 90°] (or [−π/2, π/2]). This interval covers every possible sine value from −1 to 1 exactly once.
When do you use inverse sine?
You use inverse sine whenever you know a side ratio in a right triangle and need to find an angle. For example, if the side opposite an angle is 5 and the hypotenuse is 10, you compute sin⁻¹(5/10) = sin⁻¹(0.5) = 30°. Inverse sine also appears in calculus, physics (projectile motion, wave equations), and solving trigonometric equations.

Inverse Sine (arcsin) vs. Inverse Cosine (arccos)

Inverse Sine (arcsin)Inverse Cosine (arccos)
Notationsin⁻¹(x) or arcsin(x)cos⁻¹(x) or arccos(x)
Domain (input)[−1, 1][−1, 1]
Range (output)[−π/2, π/2] or [−90°, 90°][0, π] or [0°, 180°]
Returns 0 when input is0 (since sin 0 = 0)1 (since cos 0 = 1)
Negative input givesA negative angleAn angle between 90° and 180°
Relationshiparcsin(x) + arccos(x) = π/2arccos(x) + arcsin(x) = π/2

Why It Matters

Inverse sine is essential in trigonometry courses whenever you solve for an unknown angle in a right triangle or use the Law of Sines. It appears repeatedly in physics—for example, calculating the launch angle needed for a projectile or the angle of refraction in Snell's law. Understanding its restricted range prevents errors when solving trigonometric equations that have multiple solutions.

Common Mistakes

Mistake: Confusing sin⁻¹(x) with 1/sin(x).
Correction: The notation sin⁻¹(x) means the inverse sine function, not the reciprocal. The reciprocal of sine is cosecant: 1/sin(x) = csc(x). The superscript −1 here denotes an inverse function, not an exponent.
Mistake: Choosing an angle outside the range [−90°, 90°].
Correction: Multiple angles share the same sine value (e.g., sin 30° = sin 150° = 0.5). The inverse sine function always returns the angle in [−90°, 90°]. If a problem asks for all solutions, you must use the general solution formulas separately—sin⁻¹ alone gives only the principal value.

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