For sin(θ) = -0.4321, 0 ≤ θ ≤ 360°: Identities and Equations http://postimage.org/image/dbp97g807/
Determine the following:
a. How many solutions are possible?
>> A sine function has a positive value if angle `theta` is at first and second quadrant of a unit circle. And it will have a negative value if `theta` is at third and fourth quadrant.
>>Hence, `sin theta = -0.4321` has two possible solutions at the interval `0lt=thetalt=360^o` .
b. In which quadrants would you find the solutions.
>> In sin `theta` = -0.4321, the angle `theta` is at the third and fourth quadrant of a unit circle.
c. Determine the reference angle to the nearest degree. (Note that in calculator, if we compute the angle, the value is always an acute angle, either positive or negative.)
>> `sin theta = -0.4321`
`theta_(ref) = sin^-1 -0.4321`
`theta_(ref) = -26^o`
Then, take the absolute value of theta.
Hence, the reference angle is ` 26^o` .
d. Determine all the solutions to the equation to the nearest degree.
The reference angle of the equation is `26^o` . This is the angle measured from the x-axis. As mentioned above, `theta` is located at the third and fourth quadrant of the unit circle.
At third quadrant,
`theta` = 180 + reference angle `= 180 + 26 = 206^o`
At fourth quadrant,
`theta` = 360 - reference angle `= 360 - 26 = 334^o`
Thus, at the interval `0lt=thetalt=360` , the solutions are:
`theta= 206^o` and `theta = 334^o`