If sin(z) = 0.3944 and pi/2 < z < 3pi/2, what is the value of z to the nearest ten-thousandth?
To solve this problem, we need to find the value of z within the given range where sin(z) is approximately equal to 0.3944.
The first step is to use the inverse sine function, also known as arcsin or sin^(-1), to solve for z. We have sin(z) = 0.3944, so we can write that as z = arcsin(0.3944). However, be aware that arcsin only gives the principal value, which means it gives one solution within the range -pi/2 to pi/2.
Since the given range is pi/2 < z < 3pi/2, we need to find the value of z within this specified range. Let's evaluate arcsin(0.3944) to get the principal value:
z = arcsin(0.3944)
z ≈ 0.4020
However, this result is outside the desired range. To find the value within the range pi/2 < z < 3pi/2, we need to use the property of the sine function.
The sine function has a periodicity of 2pi, meaning that sin(z) = sin(z + 2kpi) for any integer k. Therefore, if z is outside the range -pi/2 to pi/2, we can find an equivalent angle within the range by subtracting or adding integer multiples of 2pi.
In this case, to find an equivalent angle within the range pi/2 < z < 3pi/2, we can subtract 2pi from 0.4020:
z = 0.4020 - 2pi
z ≈ -1.7405
Therefore, to the nearest ten-thousandth, the value of z within the specified range is approximately -1.7405.