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To give some background to the question, I am preparing for an admissions test for a physics course at university (the PAT), and for this test we are not allowed to use calculators or tables in any form.

In the maths section of the test, there are often trigonmetric equations such as the one below which we have to solve for $x$ within a certain range (usually $[0,\pi]$):

Solve $\sin{3x}=\sqrt{3}\cos{3x}$, for $x$ in the range $0\leq x \leq \pi$.

This problem is easy to reduce:

$$\tan{3x}=\sqrt{3}\implies3x=\tan^{-1}{\sqrt{3}}\implies x=\frac{\tan^{-1}{\sqrt{3}}}{3}$$

However, I am unsure how I can evaluate expressions such as $\tan^{-1}{\sqrt{3}}$ without a calculator, is there any method I can use for expressions like these, or am I required to simply learn various common values of $\tan$ (i.e. $\tan{\pi}$, $\tan{\frac{\pi}{4}}$, $\tan{0}$, etc.)?

Thanks in advance!

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  • $\begingroup$ use 30;60;90 degree triangles or 90;45;45 degree triangles. $\endgroup$ – fosho Sep 25 '12 at 18:39
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Typically for those exams if you are asked to find an angle it will be a common one like $\pi/2$, $\pi/3$ $\pi/6$ or integer multiples of those. A lot of times you should try and look for familiar expressions that might be part of a special right triangle. In the case of $\arctan{\sqrt{3}}$, the $\sqrt{3}$ is the important part. This tells us $\tan{\theta} = \frac{\sqrt{3}}{1}$. That means the legs of our right triangle have ratio $1:\sqrt{3}:2$. We know such a a triangle has angles $30,60,90$.

Alternatively you can use the taylor expansion of $\arctan$. You know that $\arctan{x} = \int{\frac{1}{1+x^2}}$. Now, $$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots.$$ Therefore, $$\arctan{x} = x - \frac{x^{3}}{3} + \frac{x^5}{5} - \cdots$$ If you were asked something like "find $\arctan(1.5)$" you could plug this into the expansion and get a good approximation.

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  • $\begingroup$ Thank you, I have found that we are supposed to memorize key values of $\sin$, $\cos$ and $\tan$ after all. And yeah, I know that the taylor series expansion could be used, but thanks anyway! $\endgroup$ – Thomas Russell Sep 25 '12 at 19:17
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The given equation is equavalent to \begin{equation*} \tan 3x = \sqrt{3} \Leftrightarrow 3x = \dfrac{\pi}{3} + k\pi \Leftrightarrow x = \dfrac{\pi}{9} + \dfrac{k\pi}{3}, \quad k \in \mathbb{Z}. \end{equation*} Because $0 \leqslant x \leqslant \pi$, therefore $ 0 \leqslant \dfrac{\pi}{9} + \dfrac{k\pi}{3} \leqslant \pi$, $k$ is a integer number, then $k = 0$, $k = 1$ and $k = 2$. Thus, the roots of the given equation are $x = \dfrac{\pi}{9}$, $x = \dfrac{4\pi}{9}$ and $x = \dfrac{7\pi}{9}$.

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