Trig homework: Please help!

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yerin00kim

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Feb 6, 2012
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im having a lot of trouble with this topic so please try to help!
Tan theta= cot 5 theta
SOLVE FOR THETA.

thank you! :D
 
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What have you tried? Did you consider the reciprocal relationship? tan(x) = 1/cot(x)?
 

\(\displaystyle \text{Is that }\cot(5\theta)\,\text{ or }\,\cot^5\!\theta\,?\)
 
yes i tried that and i also tried tan Θ= sinΘ /cosΘ.
here's what i did: tanΘ=cot5Θ
sinΘ= cos5Θ
 cosΘ sinΘ
but im not sure.. can i do this?
 
Hello, yerin00kim!

Recall this identity: .\(\displaystyle \cos(A+B) \:=\:\cos A\cos B - \sin A\sin B\)

\(\displaystyle \text{Solve for }\theta\!:\;\;\tan\theta \:=\:\cot5\theta\)
Click to expand...

We have: .\(\displaystyle \displaystyle \tan\theta \:=\:\cot5\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta} \:=\:\frac{\cos5\theta}{\sin5\theta} \quad\Rightarrow\quad \cos5\theta\cos\theta \:=\:\sin5\theta\sin\theta \)

. . . . . . . . \(\displaystyle \cos5\theta\cos\theta - \sin5\theta\sin\theta \;=\;0\)

. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \cos6\theta \;=\;0\)

. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle 6\theta \;=\;\frac{\pi}{2} + \pi n \)

. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \theta \;=\;\frac{\pi}{12} + \frac{\pi}{6}n \)
 
\(\displaystyle tan\theta \ =\ cot5\theta \)

\(\displaystyle tan\theta \ =\tan(\frac{\pi}{2} \ - \ 5\theta) \)

\(\displaystyle \theta \ =\ \frac{\pi}{2} \ - \ 5\theta \ \pm n\pi \)

\(\displaystyle \theta \ =\ \frac{\pi}{12} \ \ \pm n\frac{\pi}{6} \)
 
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