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

[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

 

[MarkFL]

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:

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

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

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

At this point, we may assume:

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

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

And we must have:

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

Or:

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

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

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

Hence:

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

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

 

[Elissa89]

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

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

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

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

At this point, we may assume:

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

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

And we must have:

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

Or:

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

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

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

Hence:

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

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

Thank you! Looks like I had the right idea.