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Trigonometry cot(θ) = tan(2θ - 3π) find 0 < θ < 2π

karush

Well-known member
Jan 31, 2012
2,779
$\cot{(\theta)}=\tan{(2\theta-3\pi)}$ find $0<\theta<2\pi$

From the periodic Formula $\tan{(\theta+\pi n)}=\tan{\theta}$

thus
$
\displaystyle
\cot{(\theta)}
=\tan{(2\theta)}
\Rightarrow
\frac{1}{\tan{\theta}}
=\tan{(2\theta)}
$

there are 6 answers to this, but stuck here
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would stop at:

\(\displaystyle \cot(\theta)=\tan(2\theta)\)

Then, I would try combining the following:

Co-function identity:

\(\displaystyle \cot(\theta)=\tan\left(\frac{\pi}{2}-\theta \right)\)

Periodicity of tangent function:

\(\displaystyle \tan(\theta)=\tan(\theta+k\pi)\) where \(\displaystyle k\in\mathbb{Z}\)

This will give you the six roots you desire.
 

karush

Well-known member
Jan 31, 2012
2,779
by this I assume

$\displaystyle 2 \theta =\frac{\pi}{2}-\theta$
so
$\displaystyle
\theta=\frac{\pi}{6}$
then for $\displaystyle 0<\theta<2\pi $ using $\tan{(\theta+\pi n)}=\tan{\theta}$ for period
$$
\theta = \frac{\pi}{6}, \frac{7\pi}{6}
$$
or
$$30^o,210^o$$

but that is only 2 of them
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You would actually have:

\(\displaystyle \theta=\frac{\pi}{2}-2\theta+k\pi\)

See what you get from that. :D
 

soroban

Well-known member
Feb 2, 2012
409
Hello, karush!

[tex]\cot \theta\:=\:\tan(2\theta-3\pi),\quad 0<\theta<2\pi[/tex]

[tex]\tan(2\theta - 3\pi) \;=\;\frac{\tan(2\theta) - \tan(3\pi)}{1 + \tan(2\theta)\tan(3\pi)} \;=\;\tan(2\theta)[/tex]

. . . . . . . . . . [tex]=\;\frac{2\tan\theta}{1-\tan^2\theta}[/tex]

The equation becomes: .[tex]\frac{1}{\tan\theta} \;=\;\frac{2\tan\theta}{1-\tan^2\theta}[/tex]

. . [tex]1 - \tan^2\theta \:=\:2\tan^2\theta \quad\Rightarrow\quad 3\tan^2\theta \:=\:1[/tex]

. . [tex]\tan^2\theta \:=\:\frac{1}{3} \quad\Rightarrow\quad \tan\theta \:=\:\pm\frac{1}{\sqrt{3}}[/tex]

Therefore: .[tex]\theta \;=\;\frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6}[/tex]