Trig homework: Please help!

[yerin00kim]

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

thank you!

 

[tkhunny]

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

 

[soroban]

$\displaystyle \text{Is that }\cot(5\theta)\,\text{ or }\,\cot^5\!\theta\,?$

 

[Deleted member 4993]

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

thank you!

Hint:

cot(Θ) = tan (π/2-Θ)

 

[yerin00kim]

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?

 

[soroban]

Hello, yerin00kim!

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

$\displaystyle \text{Solve for }\theta\!:\;\;\tan\theta \:=\:\cot5\theta$

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

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

. . . . . . . . . . . . . . . . . . . . . .$\displaystyle \cos6\theta \;=\;0$

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

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \theta \;=\;\frac{\pi}{12} + \frac{\pi}{6}n$

 

[Deleted member 4993]

$\displaystyle tan\theta \ =\ cot5\theta$

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

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

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