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

#### Elissa89

##### Member
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

#### MarkFL

Staff member
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:

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

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

#### Elissa89

##### Member
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.