Exercise 45 Page 925

We are given the trigonometric equation sinθ= 12. We want to describe a method to solve this equation in the interval between 0^(∘) and 360^(∘). We can first use the inverse sine function to isolate θ on the left-hand side. sinθ=1/2 ⇒ θ=sin^(- 1)( 1/2 ) We are looking for the angles whose sine value is 12. Here we will present two different methods to find them.

Using Trigonometric Ratios for a Right Triangle

Using trigonometric ratios for a right triangle, we can determine the values of θ that fit the given equation. Recall that, for an acute angle θ of a right triangle, the sine of θ is the ratio between the lengths of the opposite side and the hypotenuse. sinθ= Opposite Hypotenuse ⇔ sinθ= 1 2We can draw a right triangle showing this relationship.

For this right triangle, the leg corresponding to the angle θ is half the hypotenuse. Therefore, it is a 30^(∘)-60^(∘)-90^(∘) triangle, and the value of θ is 30^(∘). 30^(∘)=sin^(- 1)( 1/2 ) Since sine is positive in Quadrants I and II, and because a half turn measures 180^(∘), we can subtract 30^(∘) from 180^(∘) to get the measure of another angle in Quadrant II that will solve the given equation. 180^(∘) - 30^(∘)= 150^(∘) Using this method, we found that 30^(∘) and 150^(∘) are solutions to the equation θ=sin^(- 1)( 1/2 ) in the interval from 0^(∘) to 180^(∘). 30^(∘)=sin^(- 1)( 1/2 ) or 150^(∘)=sin^(- 1)( 1/2 )

Using a Unit Circle

We can also find the values of θ by using a unit circle. Recall that the y-coordinate of any point on a unit circle represents its sine value. Therefore, we first need to consider the points that have a y-coordinate of 12. Remember, sine is positive in Quadrants I and II.

Note that the points form right triangles each with a leg on the x-axis. The length of the other leg of each triangle is 12 since the y-coordinate of both points is 12. Moreover, since the length of the radius is 1 in a unit circle, the hypotenuse of each triangle is 1.

Next, we will examine the right triangles to find the angles. For both of them, one of the legs is half the hypotenuse. Therefore, they are 30^(∘)-60^(∘)-90^(∘) triangles. In this type of triangle, the smaller angle measures 30^(∘). Similar to the previous method, we we will subtract 30^(∘) from 180^(∘) to find the angle in Quadrant II.

We found that 30^(∘) and 150^(∘) are solutions to the equation θ=sin^(- 1)( 1/2 ) in the interval from 0^(∘) to 180^(∘). 30^(∘)=sin^(- 1)( 1/2 ) or 150^(∘)=sin^(- 1)( 1/2 )