# For sin(θ) = -0.4321, 0 ≤ θ ≤ 360°: Identities and Equations http://postimage.org/image/dbp97g807/

### 1 Answer | Add Yours

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`