# Evaluate cos 2(theta) and sin 2(theta)

• MHB
• Elissa89
In summary: So in summary, using the given value for tangent, we can use the double-angle identity to find the values of cosine and sine for 2(theta), which are -7/9 and -4[sqrt(2)]/9 respectively.
Elissa89
So my math professor gave us a study guide for the final but he's not aloud to give us the answers so I have no idea if my answers are correct or not. So if a few people could let me know what they got after trying this that would be great.

If tan(theta) = -2[sqrt(2)], and theta is between 270 degrees and 360 degrees, evaluate cos 2(theta) and sin 2(theta).

I got -7/9

Elissa89 said:
So my math professor gave us a study guide for the final but he's not aloud to give us the answers so I have no idea if my answers are correct or not. So if a few people could let me know what they got after trying this that would be great.

If tan(theta) = -2[sqrt(2)], and theta is between 270 degrees and 360 degrees, evaluate cos 2(theta) and sin 2(theta).

I got -7/9

I would begin with the double-angle identity for the tangent function:

$$\displaystyle \tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$$

Using the value given for $$\tan(\theta)$$, we then have:

$$\displaystyle \tan(2\theta)=\frac{2(-2\sqrt{2})}{1-(-2\sqrt{2})^2}=\frac{4\sqrt{2}}{8-1}=\frac{4\sqrt{2}}{7}$$

At this point, we may assume:

$$\displaystyle \sin(2\theta)=4\sqrt{2}r$$

$$\displaystyle \cos(2\theta)=7r$$

And we must have:

$$\displaystyle \sin^2(2\theta)+\cos^2(2\theta)=1$$

Or:

$$\displaystyle 32r^2+49r^2=1\implies r^2=\frac{1}{81}$$

Given that $$\displaystyle \sin(2\theta)<0$$, we then conclude:

$$\displaystyle r=-\frac{1}{9}$$

Hence:

$$\displaystyle \sin(2\theta)=-\frac{4\sqrt{2}}{9}$$

$$\displaystyle \cos(2\theta)=-\frac{7}{9}\quad\checkmark$$

MarkFL said:
I would begin with the double-angle identity for the tangent function:

$$\displaystyle \tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$$

Using the value given for $$\tan(\theta)$$, we then have:

$$\displaystyle \tan(2\theta)=\frac{2(-2\sqrt{2})}{1-(-2\sqrt{2})^2}=\frac{4\sqrt{2}}{8-1}=\frac{4\sqrt{2}}{7}$$

At this point, we may assume:

$$\displaystyle \sin(2\theta)=4\sqrt{2}r$$

$$\displaystyle \cos(2\theta)=7r$$

And we must have:

$$\displaystyle \sin^2(2\theta)+\cos^2(2\theta)=1$$

Or:

$$\displaystyle 32r^2+49r^2=1\implies r^2=\frac{1}{81}$$

Given that $$\displaystyle \sin(2\theta)<0$$, we then conclude:

$$\displaystyle r=-\frac{1}{9}$$

Hence:

$$\displaystyle \sin(2\theta)=-\frac{4\sqrt{2}}{9}$$

$$\displaystyle \cos(2\theta)=-\frac{7}{9}\quad\checkmark$$

Thank you! Looks like I had the right idea.

## 1. What is the formula for evaluating cos 2(theta) and sin 2(theta)?

The formula for evaluating cos 2(theta) is cos^2(theta) - sin^2(theta). For sin 2(theta), the formula is 2sin(theta)cos(theta).

## 2. How do you find the value of cos 2(theta) and sin 2(theta)?

To find the value of cos 2(theta), you can use the double angle identity cos 2(theta) = cos^2(theta) - sin^2(theta). Similarly, for sin 2(theta), you can use the double angle identity sin 2(theta) = 2sin(theta)cos(theta).

## 3. Can you provide an example of evaluating cos 2(theta) and sin 2(theta)?

For example, if theta = 30 degrees, then cos 2(30) = cos^2(30) - sin^2(30) = (cos(30))^2 - (sin(30))^2 = (sqrt(3)/2)^2 - (1/2)^2 = 3/4 - 1/4 = 1/2. Similarly, sin 2(30) = 2sin(30)cos(30) = 2(1/2)(sqrt(3)/2) = sqrt(3)/2.

## 4. How does the value of cos 2(theta) and sin 2(theta) change with different values of theta?

The values of cos 2(theta) and sin 2(theta) will change as theta changes. As theta increases or decreases, the values of cos 2(theta) and sin 2(theta) will also increase or decrease accordingly.

## 5. How are cos 2(theta) and sin 2(theta) related to each other?

Cos 2(theta) and sin 2(theta) are related through the double angle identity sin 2(theta) = 2sin(theta)cos(theta). This shows that the values of cos 2(theta) and sin 2(theta) are dependent on each other and can be calculated using this relationship.

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