Using -pi/2 ≤ y ≤ pi/2, what are the following solutions?
y=arcsin root3/2
y=arcsin 1/2
for both cases, since for any -π/2 < y < 0 the sine will be negative, we only have to concern ourself with angles in the first quadrant.
so if y = arcsin (√3/2)
then y = π/3 ( I know sin 60° = sin π/3 = √3/2 )
and for the second one
y = π/6
To find the solutions to these equations, we will use the properties of the inverse sine function or arcsine. The inverse sine function gives us the angle whose sine is equal to a given number.
1. y = arcsin(sqrt(3)/2):
First, we need to find the values of y within the given range of -π/2 ≤ y ≤ π/2 that satisfy this equation.
The inverse sine of sqrt(3)/2 is the angle whose sine is sqrt(3)/2. One such angle is π/3 radians or 60 degrees. However, π/3 is outside the range of -π/2 to π/2.
To find the angle within the given range, we need to determine which quadrant the angle is in. Since sine is positive in the first and second quadrants, we look for the angle with a sine value of sqrt(3)/2 within -π/2 to π/2.
In the first quadrant, the angle π/3 satisfies sine(π/3) = sqrt(3)/2. Therefore, π/3 is a solution.
In the second quadrant, the angle π - π/3 also satisfies sine(π - π/3) = sqrt(3)/2. Therefore, π - π/3 is another solution.
Thus, the solutions to y = arcsin(sqrt(3)/2) within the given range are y = π/3 and y = π - π/3.
2. y = arcsin(1/2):
Similar to the previous equation, we look for the angle within the given range where the sine is equal to 1/2.
In the first quadrant, the angle π/6 satisfies sine(π/6) = 1/2. Therefore, π/6 is a solution.
In the second quadrant, the angle π - π/6 also satisfies sine(π - π/6) = 1/2. Therefore, π - π/6 is another solution.
Thus, the solutions to y = arcsin(1/2) within the given range are y = π/6 and y = π - π/6.