Inverse Sine: Evaluate & Interpret (Principal Values)

1) 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), 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), the correct value is π/6.

2) 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

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].

3) 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

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.

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

−1

0

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.

5) 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°.

6) If sin(θ) = 0.6, find θ = arcsin(0.6) to the nearest tenth of a degree.

36.9°

53.1°

30.0°

60.0°

Using a calculator, arcsin(0.6) ≈ 36.87°. Rounded to the nearest tenth, θ = 36.9°. This value lies within the principal range for arcsin.

Explanation

Using a calculator, arcsin(0.6) ≈ 36.87°. Rounded to the nearest tenth, θ = 36.9°. This value lies within the principal range for arcsin.

7) 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.

8) 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.

9) Evaluate arcsin(−1) in radians.

0

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.

10) 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.

11) What angle in radians has a sine value of √2/2 within the principal range of arcsin?

−π/4

0

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.

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

−π/2

−π/4

0

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.

13) Solve for x (principal value): sin(x) = 0.95.

X = arcsin(0.95) and x = π − arcsin(0.95)

X = arcsin(0.95) only (principal)

X = −arcsin(0.95) only (principal)

No solution

The principal value of arcsin is always between −π/2 and π/2. So the first solution, arcsin(0.95), is the principal one. The second value (π − arcsin(0.95)) is valid for sine equations but lies outside the arcsin range.

Explanation

The principal value of arcsin is always between −π/2 and π/2. So the first solution, arcsin(0.95), is the principal one. The second value (π − arcsin(0.95)) is valid for sine equations but lies outside the arcsin range.

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

0.64

0.64π

0.35

0.52

arcsin(3/5) = arcsin(0.6). On a calculator, arcsin(0.6) ≈ 0.6435 radians. Rounded to two decimals, it’s 0.64 radians.

Explanation

arcsin(3/5) = arcsin(0.6). On a calculator, arcsin(0.6) ≈ 0.6435 radians. Rounded to two decimals, it’s 0.64 radians.

15) 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

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.

16) 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.

17) 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.

18) If arcsin(a) = π/6, find a.

1/2

√3/2

√2/2

1

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.

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).