Exercise 41 Page 963

We are asked to choose an angle A and find its sine and cosine values. Then, we will calculate cos 2A and sin A2 by using trigonometric identities.

Finding sin A and cos A

Let the angle measure A be 60^(∘). We will first calculate sin A and cos A using the unit circle. Recall that the length of the radius is 1 in a unit circle.

The length of the leg on the x-axis is the x-coordinate of P, which is cos 60^(∘). The length of the other leg is the y-coordinate of P, which is sin 60^(∘). Keep in mind that P is in Quadrant I. Therefore, the trigonometric ratios sin 60^(∘) and cos 60^(∘) are both positive. P(x,y) = P(cos 60^(∘), sin 60^(∘))We will now focus on the right triangle. Notice that the length of the hypotenuse is equal to the length of the radius. Therefore, the triangle has a hypotenuse that measures 1.

In a 30^(∘)-60^(∘)-90^(∘) triangle, the shorter leg is half of the hypotenuse, and the longer leg is sqrt(3) times the shorter leg. Using this, we can calculate sin 60^(∘) and cos 60^(∘).

Let's now write the values of sin 60^(∘) and cos 60^(∘). &sin 60^(∘)=sqrt(3)/2 [1em] &cos 60^(∘)= 1/2

Finding cos 2A

We will calculate cos 2A = cos 120^(∘). To do so, let's use one of the Double Angle Identities for cosine. cos 2A =cos^2 A - sin^2 A We will use this identity for A = 60^(∘) by substituting sin 60^(∘)= sqrt(3)2 and cos 60^(∘)= 12 and solving it for cos 2A=cos 120^(∘).

Finding sin A2

Finally, we will calculate sin A2 = sin 30^(∘). To do this, we will use the Half Angle Identity for sine. sin A/2 = ±sqrt(1 - cos A/2) Now, we will use this identity for A = 60^(∘) by substituting cos 60^(∘)= 12 and solving it for sin A2=sin 30^(∘).

Since the angle 30^(∘) is in Quadrant I, the sign of sin 30^(∘) is positive. Therefore, sin 30^(∘) = 12.

Examples

We can also consider other angle measures for A and calculate all of the same values using the different angles. Let's take a look at some examples.