we used sin⁻¹ to find the angle whose sine is 0.5.
我们用 sin⁻¹ 找到正弦为 0.5 的角度。
the calculator showed the result of sin⁻¹(√3/2) was π/3.
计算器显示 sin⁻¹(√3/2) 的结果是 π/3。
to solve for θ, we applied the sin⁻¹ function to both sides.
为了解出 θ，我们对两边都应用了 sin⁻¹ 函数。
the range of sin⁻¹ is restricted to [-π/2, π/2] for accurate results.
sin⁻¹ 的范围限制在 [-π/2, π/2] 以获得准确的结果。
we need to ensure the argument of sin⁻¹ is within the domain.
我们需要确保 sin⁻¹ 的参数在域内。
the inverse sine, or sin⁻¹, is a trigonometric function.
反正弦，或 sin⁻¹，是一种三角函数。
using sin⁻¹ on -1 gives us an angle of -π/2.
对 -1 使用 sin⁻¹ 得到一个角度为 -π/2。
the problem required finding the angle whose sine was 1/2 using sin⁻¹.
问题要求使用 sin⁻¹ 找到正弦为 1/2 的角度。
we can find the reference angle using sin⁻¹ and the quadrant.
我们可以使用 sin⁻¹ 和象限来找到参考角。
the sin⁻¹ of 0 is 0 radians.
0 的 sin⁻¹ 是 0 弧度。
the value of sin⁻¹(0.866) is approximately π/3.
sin⁻¹(0.866) 的值约为 π/3。
we used sin⁻¹ to find the angle whose sine is 0.5.
我们用 sin⁻¹ 找到正弦为 0.5 的角度。
the calculator showed the result of sin⁻¹(√3/2) was π/3.
计算器显示 sin⁻¹(√3/2) 的结果是 π/3。
to solve for θ, we applied the sin⁻¹ function to both sides.
为了解出 θ，我们对两边都应用了 sin⁻¹ 函数。
the range of sin⁻¹ is restricted to [-π/2, π/2] for accurate results.
sin⁻¹ 的范围限制在 [-π/2, π/2] 以获得准确的结果。
we need to ensure the argument of sin⁻¹ is within the domain.
我们需要确保 sin⁻¹ 的参数在域内。
the inverse sine, or sin⁻¹, is a trigonometric function.
反正弦，或 sin⁻¹，是一种三角函数。
using sin⁻¹ on -1 gives us an angle of -π/2.
对 -1 使用 sin⁻¹ 得到一个角度为 -π/2。
the problem required finding the angle whose sine was 1/2 using sin⁻¹.
问题要求使用 sin⁻¹ 找到正弦为 1/2 的角度。
we can find the reference angle using sin⁻¹ and the quadrant.
我们可以使用 sin⁻¹ 和象限来找到参考角。
the sin⁻¹ of 0 is 0 radians.
0 的 sin⁻¹ 是 0 弧度。
the value of sin⁻¹(0.866) is approximately π/3.
sin⁻¹(0.866) 的值约为 π/3。
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