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

Elissa89

Member
Oct 19, 2017
52
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

Administrator
Staff member
Feb 24, 2012
13,736
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
Oct 19, 2017
52
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.