- Thread starter
- #1

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

- Thread starter Elissa89
- Start date

- Thread starter
- #1

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

- Admin
- #2

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

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

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

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:

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

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

- Thread starter
- #3

Thank you! Looks like I had the right idea.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:

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

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