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^(∘).
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^(∘).
