Draw a unit circle and mark each point on the circle that has an x-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.
30^(∘) +360^(∘) * n and 330^(∘) +360^(∘) * n, for any integer n
Practice makes perfect
We want to find all the angles whose cosine is sqrt(3)2. Since we know the value of cosine but not the value of the angle, the process of finding the missing value uses an inverse trigonometric function.
cos ^(- 1) ( sqrt(3)2)
First, we need to draw the angle in standard position on a unit circle. The cosine of an angle in standard position is the x-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 that is not 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 shorter leg is 30^(∘). Let's add this information to our coordinate plane.
Now, considering that a full turn is 360^(∘), we can find the angles in standard position whose cosine is sqrt(3)2. To do so we will add 30 to 0 and subtract 30 from 360.
0+ 30&= 30^(∘)
360- 30&= 330^(∘)
Let's show the obtained angles on the coordinate plane.
We found that cos^(- 1)( sqrt(3)2) can be either 30^(∘) or 330^(∘). 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 cosine of the resulting angles will also be sqrt(3)2.
30^(∘)+360^(∘) * n and 330^(∘)+360^(∘) * n,
wheren is any integer
