A problem with cotangent squared: 1.6666 = cot^2 (theta/2)

Vix

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Apr 3, 2007
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Okay, I'm unable to solve this with my calculator. I need to modify the equation so I have theta isolated on one side of the equation. I don't really know how to solve it on paper either. I'm really not sure how cotangent squared works.

1.6666 = cot^2 (theta/2)

Thanks so much!!
 
\(\displaystyle \L\\\frac{5}{3}=cot^{2}(\frac{{\theta}}{2})\)

Square root of both sides:

\(\displaystyle \L\\\sqrt{\frac{5}{3}}=cot(\frac{{\theta}}{2})\)

Arccot:

\(\displaystyle \L\\cot^{-1}(\sqrt{\frac{5}{3}})=\frac{{\theta}}{2}\)

\(\displaystyle \L\\2cot^{-1}(\sqrt{\frac{5}{3}})={\theta}\)
 
Re: A problem with cotangent squared

Hello, Vix!

Since we don't have a \(\displaystyle \fbox{\cot}\) key on our calculator,
. . I would solve for \(\displaystyle \theta\) like this . . .


We have: \(\displaystyle \L\:\frac{5}{3} \;=\;\cot^2\left(\frac{\theta}{2}\right)\)

. . . . . . . . \(\displaystyle \L\frac{5}{3}\;=\;\frac{1}{\tan^2\left(\frac{\theta}{2}\right)}\)

\(\displaystyle \L\frac{5}{3}\cdot\tan^2\left(\frac{\theta}{2}\right) \;=\;1\)

.. \(\displaystyle \L\tan^2\left(\frac{\theta}{2}\right) \;=\;\frac{3}{5} \;=\;0.6\)

. . \(\displaystyle \L\tan\left(\frac{\theta}{2}\right) \;=\;\pm\sqrt{0.6}\)

. . . . . . . .\(\displaystyle \L\frac{\theta}{2}\;=\;\tan^{-1}\left(\pm\sqrt{0.6}\right)\)

. . . . . . . .\(\displaystyle \L\theta \;= \;2\cdot\tan^{-1}\left(\pm\sqrt{0.6}\right)\)

 
Thank You!

Thanks so much. I really appreciate the speedy reply!
 
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