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If $\tan^23\theta = 1$, how do I manipulate the equation so I can make $\tan\theta$ the subject? I forgot how to do these since it has been a long time. I tried searching before posting. My answer is $\frac{1}{\sqrt3}$, not sure if it is correct though.

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closed as off-topic by alexqwx, Antonio Vargas, Semiclassical, Claude Leibovici, PVAL Aug 14 '14 at 5:04

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up vote 6 down vote accepted

$$\tan^2(3\theta) = 1 \iff \sqrt{\tan^2(3\theta)} = \pm \sqrt 1 \iff \tan(3\theta) = \pm 1$$

Then $$\theta = \frac 13\tan^{-1}(1)\tag{I}$$ or $$\theta = \frac 13\tan^{-1}(-1)\tag{II}$$


  • $\theta = \frac 13(45^{\circ}) = 15^\circ$ or

  • $\theta = \frac 13(225^\circ) = 75\circ$ or

  • $\theta = \frac 13(405^\circ) = 135^\circ$ or

  • $\theta = \frac 13(585^\circ) = 195^\circ$ or

  • $\theta = \frac 13(765^\circ) = 255^\circ$ or

  • $\theta = \frac 13(945^\circ) = 315^\circ$.

    I assume you are looking for solutions in $0 \leq \theta \leq 360^\circ$.

Can you do the same for $(II)$?

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Can you tell me where you got the values 45, 225, 405, 585, 765, 945 from? – Qwerty Aug 9 '14 at 13:29
Sure, because we want all theta between 0 and 360, we need to consider three rotations around the circle: that means we want $3\theta$ between $0$ and $3\cdot 360 = 1080$ degrees. We take for I., $3\theta = 45 + k180$ until we get all solutions for $3\theta$ in that range, and solve each for $\theta = \frac 13(3\theta)$. For $II$, we need all solutions such that $3\theta = 135 + k 180$, for integers k, for $0\leq 3\theta \leq 1080^\circ$. – amWhy Aug 9 '14 at 13:37
Note that $\tan(3\theta) = 1 \implies 3\theta = 45^\circ +k180)$ – amWhy Aug 9 '14 at 13:41
Many thanks for your time! – Qwerty Aug 9 '14 at 13:43
You're welcome, Qwerty! – amWhy Aug 9 '14 at 13:45

As with any square root problem, if you have $x^2 = a$, then either $x = \sqrt{a}$ or $x = -\sqrt{a}$.

In this case, either $\tan 3\theta = 1$ or $\tan 3\theta = -1$. This should be solvable from this point.

EDIT: define $\alpha = 3\theta$, then solve for $\alpha$. Then, divide that by 3 to get $\theta$. You should get $\frac{k\pi}{12}$ for odd $k$.

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If you have $x^2=a$, then either $x=\sqrt{a}$ or $x=-\sqrt{a}$ – kingW3 Aug 9 '14 at 13:12
Herp. Good call. – Fargle Aug 9 '14 at 13:13

Other way,

$$\begin{align} \tan3\theta=\pm 1 \\ \frac{\tan \theta+\tan2\theta}{1-\tan\theta\tan2\theta}=\pm 1\\ \frac{\tan\theta-\tan^3\theta+\tan^2\theta} {1-\tan^2\theta+2\tan\theta}=\pm1 \end{align}$$

That should give us 6 results.

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Algebraically, this looks fine, but it's not very simple. We end up with a cubic equation in $\tan(\theta)$ which is usually a pain to solve. – alexqwx Aug 9 '14 at 17:09
Yes. I agree. The most elegant solution is already given above. For those who are not so good with inverse might want to go this way :) – MonK Aug 11 '14 at 8:11

Whenever you've the squares of trigonometric ratios only in one side of the equation, use one of $$\cos2x=2\cos^2x-1=1-2\sin^2x=\frac{1-\tan^2x}{1+\tan^2x}$$

So, we have $\displaystyle\cos6\theta=0\iff 6\theta=(2n+1)90^\circ\iff\theta=(2n+1)15^\circ$ where $n$ is any integer

For $\displaystyle0\le\theta<360^\circ, 0\le2n+1<24\implies0\le n\le11$ as $n$ is any integer

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@Qwerty, How about this – lab bhattacharjee Aug 9 '14 at 14:34
Thanks for that, I'll note it down in my book. – Qwerty Aug 10 '14 at 2:07

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