Solve Equations With Inverse Sine

1) Solve for x (principal value): sin(x) = 1/2.

π/3

π/4

π/6

−π/6

We know sin(π/6) = 1/2. The principal range for arcsin is [−π/2, π/2], and π/6 lies in this range. Therefore, x = arcsin(1/2) = π/6.

Explanation

We know sin(π/6) = 1/2. The principal range for arcsin is [−π/2, π/2], and π/6 lies in this range. Therefore, x = arcsin(1/2) = π/6.

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

π/6

π/4

π/3

π/2

Simplify the equation: 2sin(x) − 1 = 0 ⇒ sin(x) = 1/2. The inverse sine of 1/2 gives x = π/6. This value lies within the arcsin range.

Explanation

Simplify the equation: 2sin(x) − 1 = 0 ⇒ sin(x) = 1/2. The inverse sine of 1/2 gives x = π/6. This value lies within the arcsin range.

3) Evaluate arcsin(−√2/2) in radians.

−π/6

−π/4

−π/3

−π/2

sin(π/4) = √2/2. For a negative sine value, take the negative of the corresponding positive angle. Thus arcsin(−√2/2) = −π/4.

Explanation

sin(π/4) = √2/2. For a negative sine value, take the negative of the corresponding positive angle. Thus arcsin(−√2/2) = −π/4.

4) Which statement is true for all x in [−1, 1]?

Sin(arcsin(x)) = x

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

Arcsin(x) ∈ [0, π]

Arcsin(x) = −arccos(x)

The sine and arcsine functions are inverses. That means if x is between −1 and 1, applying sine to arcsin(x) returns x itself. So sin(arcsin(x)) = x always holds true.

Explanation

The sine and arcsine functions are inverses. That means if x is between −1 and 1, applying sine to arcsin(x) returns x itself. So sin(arcsin(x)) = x always holds true.

5) Solve for θ in [−π/2, π/2]: sin(θ) = −3/5.

θ = arcsin(−5/3)

θ = π − arcsin(3/5)

θ = −π + arcsin(3/5)

θ = −arcsin(3/5)

Because the sine value is negative, the angle must lie below the x-axis. So θ = −arcsin(3/5). This is within the principal range of arcsin.

Explanation

Because the sine value is negative, the angle must lie below the x-axis. So θ = −arcsin(3/5). This is within the principal range of arcsin.

6) Evaluate arcsin(0) in radians.

−π/2

0

π/2

π

The sine of 0 equals 0. Therefore, arcsin(0) = 0 since 0 is in the range [−π/2, π/2].

Explanation

The sine of 0 equals 0. Therefore, arcsin(0) = 0 since 0 is in the range [−π/2, π/2].

7) Solve for x (principal value): sin(x) = √3/2.

π/6

π/3

2π/3

−π/3

Since sin(π/3) = √3/2, the angle x = π/3 gives the required sine value. It’s also within the arcsin range, so x = π/3.

Explanation

Since sin(π/3) = √3/2, the angle x = π/3 gives the required sine value. It’s also within the arcsin range, so x = π/3.

8) Given sin(t) = 0.92, find the principal value t in degrees (nearest tenth).

23.6°

52.7°

66.9°

90.0°

Use a calculator to find t = arcsin(0.92). This gives t ≈ 66.9°. It lies within the arcsin range of −90° to 90°.

Explanation

Use a calculator to find t = arcsin(0.92). This gives t ≈ 66.9°. It lies within the arcsin range of −90° to 90°.

9) If arcsin(y) = −π/6, find y.

−√3/2

−1/2

√3/2

1/2

Apply sine to both sides: sin(arcsin(y)) = sin(−π/6). Since sin(−π/6) = −1/2, we get y = −1/2.

Explanation

Apply sine to both sides: sin(arcsin(y)) = sin(−π/6). Since sin(−π/6) = −1/2, we get y = −1/2.

10) Solve for x: sin(2x) = 1/2. Find the principal value for 2x, then for x.

X = π/12

X = π/6

X = π/4

X = π/3

First, 2x = arcsin(1/2) = π/6. Divide both sides by 2 → x = π/12. This gives the principal solution.

Explanation

First, 2x = arcsin(1/2) = π/6. Divide both sides by 2 → x = π/12. This gives the principal solution.

11) Which angle in radians satisfies sin(x) = 1/2 within the principal range?

−π/6

0

π/6

π/2

sin(π/6) = 1/2, and π/6 is in the arcsin range [−π/2, π/2]. Thus, the correct principal angle is π/6.

Explanation

sin(π/6) = 1/2, and π/6 is in the arcsin range [−π/2, π/2]. Thus, the correct principal angle is π/6.

12) Which angle satisfies sin(x) = −1 within the principal range?

−π/2

−π/6

0

π/2

The sine of −π/2 equals −1. Because −π/2 is included in the range [−π/2, π/2], arcsin(−1) = −π/2.

Explanation

The sine of −π/2 equals −1. Because −π/2 is included in the range [−π/2, π/2], arcsin(−1) = −π/2.

13) Solve for x (principal value): sin(x) = −0.8.

X = −arcsin(0.8)

X = π − arcsin(0.8)

X = arcsin(−1.25)

No solution

Because the sine value is negative, the angle must be negative. The principal solution is x = −arcsin(0.8). This falls within the valid arcsin range.

Explanation

Because the sine value is negative, the angle must be negative. The principal solution is x = −arcsin(0.8). This falls within the valid arcsin range.

14) Evaluate sin(arcsin(−7/10)).

−3/10

−7/10

7/10

3/10

Sine and arcsine are inverse operations. So sin(arcsin(x)) = x. Therefore, sin(arcsin(−7/10)) = −7/10.

Explanation

Sine and arcsine are inverse operations. So sin(arcsin(x)) = x. Therefore, sin(arcsin(−7/10)) = −7/10.

15) Solve 3sin(x) + 1 = 0 for the principal value of x.

X = −arcsin(1/3)

X = arcsin(−1/3)

X = −arcsin(3)

No solution

Simplify: 3sin(x) + 1 = 0 → sin(x) = −1/3. Now, take arcsin: x = arcsin(−1/3). This is within the valid range because −1 ≤ −1/3 ≤ 1.

Explanation

Simplify: 3sin(x) + 1 = 0 → sin(x) = −1/3. Now, take arcsin: x = arcsin(−1/3). This is within the valid range because −1 ≤ −1/3 ≤ 1.

16) Evaluate arcsin(−1) in radians.

0

−π/2

π/2

−π

sin(−π/2) = −1. Thus, arcsin(−1) = −π/2 since that’s in the principal range.

Explanation

sin(−π/2) = −1. Thus, arcsin(−1) = −π/2 since that’s in the principal range.

17) A ramp rises 12 inches over a horizontal run of 30 inches. Let θ be the angle with the ground. Which expression gives θ?

θ = arcsin(12/42)

θ = arcsin(12/30)

θ = arcsin(30/12)

θ = π − arcsin(12/30)

The sine of an angle equals rise/hypotenuse. So θ = arcsin(12/30). This expression gives the correct relationship between rise and hypotenuse.

Explanation

The sine of an angle equals rise/hypotenuse. So θ = arcsin(12/30). This expression gives the correct relationship between rise and hypotenuse.

18) Solve for x (principal value): sin(x) = 2/3. Round to the nearest 0.01 radians.

0.73

0.73π

0.52

0.85

Using a calculator: arcsin(2/3) ≈ 0.7297 radians. Rounded to two decimals, x = 0.73 radians. This is the principal value.

Explanation

Using a calculator: arcsin(2/3) ≈ 0.7297 radians. Rounded to two decimals, x = 0.73 radians. This is the principal value.

19) Which is the correct domain and range of y = arcsin(x)?

Domain (−∞, ∞); Range [−π/2, π/2]

Domain [−1, 1]; Range [0, π]

Domain [−1, 1]; Range [−π/2, π/2]

Domain [0, 1]; Range [0, π/2]

arcsin(x) only accepts inputs between −1 and 1. The output (angle) lies between −π/2 and π/2. That’s the definition of the function’s domain and range.

Explanation

arcsin(x) only accepts inputs between −1 and 1. The output (angle) lies between −π/2 and π/2. That’s the definition of the function’s domain and range.