Draw a unit circle and mark each point on the circle that has a y-coordinate equal to - sqrt(3)2. Use those points to draw two right triangles, each with one leg on the x-axis. Then, find the missing leg using the Pythagorean Theorem and note the type of triangle.
240^(∘)+n * 360^(∘) and 300^(∘)+n * 360^(∘), for any integer n
Practice makes perfect
We want to find all angles whose sine is - sqrt(3)2. Since we know the value of sine but not the value of the angle, the process of finding the missing value uses an inverse trigonometric function.
sin^(- 1)(- sqrt(3)2)
First, we need to draw the angle in standard position on a unit circle. The sine of an angle in standard position is the y-coordinate of the point of intersection of its terminal side and the unit circle.
On the same coordinate plane, we will use those points to draw two right triangles, each with one leg on the x-axis.
For both triangles, we have that the hypotenuse is 1 and that one of the legs has a length of sqrt(3)2. We can use the Pythagorean Theorem to find the length of the leg on the x-axis.
Note that when solving the equation we only kept the principal root because b represents a side length and must be a positive number.
We see that the shorter leg on both triangles is half the length of the hypotenuse, and that the longer leg is sqrt(3) times the length of the shorter leg. Therefore, we have two 30^(∘)-60^(∘)-90^(∘) triangles. In this type of triangle, the measure of the opposite angle to the longer leg is 60^(∘). Let's add this information to our coordinate plane.
Now, considering that a half turn is 180^(∘) and that a full turn is 360^(∘), we can find the angles in standard position whose sine is - sqrt(3)2. To do so, we will add 60 to 180 and subtract 60 from 360.
180+ 60&= 240^(∘)
360- 60&= 300^(∘)
Let's show the obtained angles on the coordinate plane.
We found that sin^(- 1)(- sqrt(3)2) can be either 240^(∘) or 300^(∘). Finally, keep in mind that if we add or subtract a multiple of 360^(∘), the terminal side of the angle will be in the same position. Therefore, the sine of the resulting angles will also be - sqrt(3)2.
240^(∘)+ n * 360^(∘) and 300^(∘)+ n * 360^(∘),
wheren is any integer
