Find two values of theta that satisfy the equation. give your answers in degrees (0degrees <theta<360degrees ) and radians (0<theta<2pi)
a. sin θ= sqrrt3/2
b.sin θ= - sqrrt3/2
To find the values of theta that satisfy the equation, we need to use the inverse sine function (also known as arcsine) which is denoted as sin^(-1) or asin.
a. sin θ = sqrt(3)/2
In degrees:
To find the first value of theta, we find the angle whose sine is sqrt(3)/2. This will be 60 degrees because sin(60°) = sqrt(3)/2. So, one value of theta that satisfies the equation in degrees is 60°.
To find the second value of theta, we use the fact that the sine function is positive in the first and second quadrants. In the second quadrant, the angle that has a sine of sqrt(3)/2 is 180° - 60° = 120°. So, the second value of theta that satisfies the equation in degrees is 120°.
In radians:
To convert the degrees to radians, you can use the conversion factor that pi radians is equivalent to 180 degrees. Therefore, the first value of theta is 60° * (pi/180°) = pi/3 radians, and the second value of theta is 120° * (pi/180°) = 2pi/3 radians.
b. sin θ = -sqrt(3)/2
In degrees:
To find the first value of theta, we find the angle whose sine is -sqrt(3)/2. This will be 240 degrees because sin(240°) = -sqrt(3)/2. So, one value of theta that satisfies the equation in degrees is 240°.
To find the second value of theta, we use the fact that the sine function is negative in the third and fourth quadrants. In the third quadrant, the angle that has a sine of -sqrt(3)/2 is 180° + 60° = 240°. So, the second value of theta that satisfies the equation in degrees is also 240°.
In radians:
Using the same conversion factor, the first value of theta is 240° * (pi/180°) = 4pi/3 radians, and the second value of theta is also 240° * (pi/180°) = 4pi/3 radians.
a. To find two values of theta that satisfy the equation sin θ = sqrt(3)/2, we can use the inverse sine function.
In degrees:
θ = sin^(-1)(sqrt(3)/2)
θ = 60 degrees
However, keep in mind that the sine function has a period of 360 degrees, so we can find another solution by adding the period.
θ = 180 - θ (since sin (180-θ) = sin θ)
θ = 180 - 60
θ = 120 degrees
In radians:
θ = sin^(-1)(sqrt(3)/2)
θ = pi/3
Since the sine function has a period of 2pi radians, we can find another solution by adding the period.
θ = 2pi - θ (since sin (2pi-θ) = sin θ)
θ = 2pi - pi/3
θ = 5pi/3
Therefore, the two values of theta that satisfy the equation sin θ = sqrt(3)/2 are θ = 60 degrees and θ = 120 degrees (or θ = pi/3 and θ = 5pi/3 in radians).
b. To find two values of theta that satisfy the equation sin θ = -sqrt(3)/2, we can again use the inverse sine function.
In degrees:
θ = sin^(-1)(-sqrt(3)/2)
θ = -60 degrees
Again, because the sine function has a period of 360 degrees, we can find another solution by adding the period.
θ = 180 - θ (since sin (180-θ) = sin θ)
θ = 180 - (-60)
θ = 240 degrees
In radians:
θ = sin^(-1)(-sqrt(3)/2)
θ = -pi/3
Since the sine function has a period of 2pi radians, we can find another solution by adding the period.
θ = 2pi - θ (since sin (2pi-θ) = sin θ)
θ = 2pi - (-pi/3)
θ = 7pi/3
Therefore, the two values of theta that satisfy the equation sin θ = -sqrt(3)/2 are θ = -60 degrees and θ = 240 degrees (or θ = -pi/3 and θ = 7pi/3 in radians).
sin θ= sqrrt3/2
This is just your standard pi/3. So, pi/3 and 2pi/3
b. 4pi/3, 5pi/3