Inverse Sine: Evaluate & Interpret (Principal Values)

Grade 9th

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| Attempts: 105 | Questions: 20 | Updated: Jan 22, 2026

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1) Evaluate arcsin(√2/2) in degrees.

30°

45°

60°

90°

We know sin(45°) = √2/2. Because 45° is within the arcsin range (−90° to 90°), the answer is 45°.

Explanation

We know sin(45°) = √2/2. Because 45° is within the arcsin range (−90° to 90°), the answer is 45°.

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About This Quiz

Inverse Sine: Evaluate & Interpret (Principal Values) - Quiz

Get comfortable with arcsin by learning how to interpret, evaluate, and connect inverse sine to the unit circle. In this quiz, you’ll find exact values like π/6, π/4, and π/3, work within the principal range [−π/2,π/2][-π/2, π/2][−π/2,π/2], and determine valid domains and ranges. You’ll practice reading and writing arcsin expressions,... see moreconnecting them to real angles, and checking which values are defined — all while reinforcing that sin(arcsin x) = x only when x lies within [−1, 1].
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2) What angle in radians has a sine value of √2/2 within the principal range of arcsin?

−π/4

π/4

π/2

Since sin(π/4) = √2/2 and π/4 is between −π/2 and π/2, the corresponding arcsin value is π/4.

Explanation

Since sin(π/4) = √2/2 and π/4 is between −π/2 and π/2, the corresponding arcsin value is π/4.

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3) Evaluate sin(arcsin(−3/5)).

−4/5

−3/5

3/5

4/5

Sine and arcsine are inverse functions. When they are applied one after another, they cancel out: sin(arcsin(x)) = x. So, sin(arcsin(−3/5)) = −3/5.

Explanation

Sine and arcsine are inverse functions. When they are applied one after another, they cancel out: sin(arcsin(x)) = x. So, sin(arcsin(−3/5)) = −3/5.

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4) Evaluate arcsin(√3/2) in radians.

π/6

π/3

π/4

π/2

We know sin(π/3) = √3/2. Since π/3 lies in arcsin’s range [−π/2, π/2], the value is π/3.

Explanation

We know sin(π/3) = √3/2. Since π/3 lies in arcsin’s range [−π/2, π/2], the value is π/3.

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5) What is the range of y = arcsin(x)?

[0, π]

[−π/2, π/2]

[−π, π]

(−∞, ∞)

The arcsin function returns the angle whose sine is x. To make it a true function, arcsin is restricted to give values between −π/2 and π/2. This ensures only one angle corresponds to each sine value.

Explanation

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6) Evaluate arcsin(−1) in radians.

π/2

−π/2

−π

The sine of −π/2 equals −1. Since −π/2 lies within the arcsin range, arcsin(−1) = −π/2.

Explanation

The sine of −π/2 equals −1. Since −π/2 lies within the arcsin range, arcsin(−1) = −π/2.

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7) A right triangle has hypotenuse 10 and opposite side 4. Express θ using inverse sine.

θ = arcsin(2/5)

θ = arcsin(4/10)

θ = arcsin(5/10)

θ = arcsin(3/5)

By definition, sin(θ) = opposite/hypotenuse = 4/10. To find θ, take the inverse sine: θ = arcsin(4/10). This expresses the angle in terms of arcsin.

Explanation

By definition, sin(θ) = opposite/hypotenuse = 4/10. To find θ, take the inverse sine: θ = arcsin(4/10). This expresses the angle in terms of arcsin.

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8) Evaluate arcsin(−√2/2) in radians.

−π/6

−π/4

−π/3

−π/2

We know sin(π/4) = √2/2. To get a negative sine, the angle must be below the x-axis, so −π/4. That’s within arcsin’s range, so arcsin(−√2/2) = −π/4.

Explanation

We know sin(π/4) = √2/2. To get a negative sine, the angle must be below the x-axis, so −π/4. That’s within arcsin’s range, so arcsin(−√2/2) = −π/4.

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9) Evaluate arcsin(1/2) in radians.

π/6

π/3

π/4

π/2

arcsin(1/2) asks for the angle whose sine equals 1/2. On the unit circle, sin(π/6) = 1/2. Since π/6 lies within the principal range of arcsin (−π/2 to π/2), so the correct value is π/6.

Explanation

arcsin(1/2) asks for the angle whose sine equals 1/2. On the unit circle, sin(π/6) = 1/2. Since π/6 lies within the principal range of arcsin (−π/2 to π/2), so the correct value is π/6.

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10) If arcsin(a) = π/6, find a.

1/2

√3/2

√2/2

Take sine on both sides: sin(arcsin(a)) = sin(π/6). This gives a = 1/2 since sin(π/6) = 1/2.

Explanation

Take sine on both sides: sin(arcsin(a)) = sin(π/6). This gives a = 1/2 since sin(π/6) = 1/2.

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11) If sin(θ) = 1/2, find θ.

30°

45°

60°

90°

The sine of an angle is equal to 1/2 at 30 degrees (or π/6 radians) in the unit circle. Therefore, θ is 30°.

Explanation

The sine of an angle is equal to 1/2 at 30 degrees (or π/6 radians) in the unit circle. Therefore, θ is 30°.

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12) Solve for θ in [−π/2, π/2]: 3sin(θ) = √3.

θ = π/6

θ = π/3

θ = arcsin(√3)

θ = arcsin(√3/3)

First, isolate sine: sin(θ) = √3 / 3 = 1 / √3. Now, θ = arcsin(√3 / 3). This angle lies in the principal range since arcsin only returns values between −π/2 and π/2.

Explanation

First, isolate sine: sin(θ) = √3 / 3 = 1 / √3. Now, θ = arcsin(√3 / 3). This angle lies in the principal range since arcsin only returns values between −π/2 and π/2.

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13) Find the angle in radians whose sine value is 0 within the principal range.

−π/2

−π/4

π/4

The sine of 0 equals 0. This is directly in the middle of the principal range, so arcsin(0) = 0.

Explanation

The sine of 0 equals 0. This is directly in the middle of the principal range, so arcsin(0) = 0.

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14) Solve for x in the principal range: sin(x) = −√3/2.

−π/6

−π/3

−π/4

−2π/3

We know sin(π/3) = √3/2. Since the sine value here is negative, the angle must be below the x-axis. Sine is negative in Quadrant IV, so we take −π/3 (the reflection of π/3). That’s within the principal range [−π/2, π/2].

Explanation

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15) Find the principal value of x such that sin(x) = 0.95.

X = arcsin(0.95)

X = π − arcsin(0.95)

X = −arcsin(0.95)

No solution

The principal value for arcsin is always taken in the interval −π/2 ≤ x ≤ π/2.

Since 0.95 is within the valid range of the sine function, arcsin(0.95) gives a real number in the first quadrant.

The second solution, π − arcsin(0.95), is valid for the full solution set but does not belong to the principal-value interval.

Therefore, the correct principal value is x = arcsin(0.95).

Explanation

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16) Solve 2sin(x) − 1 = 0 for the principal value of x.

X = π/6

X = π/4

X = π/3

X = arcsin(2)

First, isolate sine: 2sin(x) − 1 = 0 ⇒ sin(x) = 1/2. The angle whose sine is 1/2 is π/6. This is within the arcsin range, so x = π/6.

Explanation

First, isolate sine: 2sin(x) − 1 = 0 ⇒ sin(x) = 1/2. The angle whose sine is 1/2 is π/6. This is within the arcsin range, so x = π/6.

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17) Evaluate arcsin(3/5) to the nearest 0.01 radians.

0.64

0.64π

0.35

0.52

The value arcsin(3/5) represents the angle (in radians) whose sine is 3/5.

Since 3/5 = 0.6, the angle must be a little larger than arcsin(1/2) = 0.52 radians and smaller than arcsin(√3/2) = 1.05 radians.

Among the choices, 0.64 is the only value that fits this range and is close to the true value of the angle.

Explanation

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18) Which identity is always true for x in [−1, 1]?

Arcsin(sin(x)) = x for all real x

Arcsin(x) + arccos(x) = π

Arcsin(x) = arctan(x)

Sin(arcsin(x)) = x

Sine and arcsine undo each other when the input is between −1 and 1. That’s why sin(arcsin(x)) = x always holds true for valid sine values. The reverse (arcsin(sin(x))) only works in the restricted domain.

Explanation

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19) A ramp rises 18 inches over a run of 36 inches. Let θ be the angle with the ground. Which expression gives θ?

θ = arcsin(1/3)

θ = arcsin(1/2)

θ = arcsin(2)

θ = arcsin(18/36)

The sine of an angle equals the ratio of the opposite side (rise) to the hypotenuse. Here, opposite = 18 and hypotenuse = 36. So θ = arcsin(18/36) = arcsin(1/2).

Explanation

The sine of an angle equals the ratio of the opposite side (rise) to the hypotenuse. Here, opposite = 18 and hypotenuse = 36. So θ = arcsin(18/36) = arcsin(1/2).

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20) Which value is not in the domain of f(x) = arcsin(x)?

−1

1.2

Sine values only exist between −1 and 1. Since arcsin(x) undoes sine, it only accepts inputs between −1 and 1. Therefore, any number outside that range, such as 1.2, is invalid.

Explanation

Sine values only exist between −1 and 1. Since arcsin(x) undoes sine, it only accepts inputs between −1 and 1. Therefore, any number outside that range, such as 1.2, is invalid.

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