# Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$

how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?

I know how to prove $\frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0},$

in this way:

$\tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$

$\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$

$\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$

$\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$

$\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$

$\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$

Could this help to prove the first one and how ?Do i just need to know that $\frac{1}{\tan\theta}=\tan(90^{\circ}-\theta)$ ?

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Next up: $\tan(\pi/24) = \tan^2(3\pi/24) \tan(5 \pi/24)$ and $\tan(\pi/30) \tan(7 \pi/30) = \tan(2\pi/30) \tan(4 \pi/30)$ – Robert Israel Jul 18 '12 at 7:03
$\tan(\pi/21) = \tan(5 \pi/42) \tan(\pi/7) \tan(3 \pi/14)$ and $\tan(\pi/54) = \tan(\pi/9) \tan(4 \pi/27) \tan(\pi/6) \tan(5 \pi/27) \tan(2 \pi/9)$. What's the general pattern? – Robert Israel Jul 18 '12 at 16:56
Ah: one of the patterns is $\tan(x) = \tan(\pi/6+x) \tan(\pi/6-x) \tan(3x)$. Try it for $x = \pi/18$, $\pi/24$, $\pi/21$. – Robert Israel Jul 18 '12 at 20:06
– lab bhattacharjee Feb 5 at 8:43

In a word, yes. You already know that (in degrees) $\tan 20\cdot\tan30=\tan10\cdot\tan50$ so

$$\tan20\cdot\tan30\cdot\tan40 = \tan10\cdot\tan50\cdot\tan40$$

$$\frac{1}{\tan40}=\tan50$$ is all you need.

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how can we conclude from $\tan 20\cdot\tan30=\tan10\cdot\tan50$ that $\tan20\cdot\tan30\cdot\tan40 = \tan10\cdot\tan50\cdot\tan40$ ? – Frank Jul 18 '12 at 1:21
Is that mean $\tan 20\cdot\tan30=\tan10\cdot\tan50$ is equal to $\tan20\cdot\tan30\cdot\tan40 = \tan10\cdot\tan50\cdot\tan40$ is equal to $\tan20\cdot\tan30\cdot\tan40 = \tan10$ ? – Frank Jul 18 '12 at 1:27
@Mohammed, all Rick has done is to multiply both sides of the equation by $\tan40$. – Gerry Myerson Jul 18 '12 at 1:35
Yes. The $\tan 40\text{ and }\tan 50$ cancel each other out. – Rick Decker Jul 18 '12 at 1:36

This may be the hard way, but you can use $$\tan\theta={e^{i\theta}-e^{- i\theta}\over i(e^{i\theta}+e^{-i\theta})}$$ to write everything in terms of $\zeta=e^{\pi i/18}$. Then $\zeta$ is a primitive 36th root of unity, and you can use that to make simplifications.

EDIT: In light of the comments, I flesh this out a bit.

We start with Euler's formula, $$e^{ix}=\cos x+i\sin x\tag1$$ We'll say a bit more about where this comes from, later. Replace $x$ everywhere with $-x$, and use $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$, to get $$e^{-ix}=\cos x-i\sin x\tag2$$ Add (1) and (2), and divide by 2 to get $$\cos x={e^{ix}+e^{-ix}\over2}\tag3$$ Subtract (2) from (1) and divide by $2i$ to get $$\sin x={e^{ix}-e^{-ix}\over2i}\tag4$$ Divide (4) by (3) to get $$\tan x={e^{ix}-e^{-ix}\over i(e^{ix}+e^{-ix})}\tag5$$ 10 degrees is $\pi/18$ radians; put in $x=\pi/18$ to get $$\tan{\pi\over18}={e^{i{\pi\over18}}-e^{-i{\pi\over18}}\over i(e^{i{\pi\over18}}+e^{-i{\pi\over18}})}\tag6$$ Let $\zeta=e^{\pi i/18}$; then $$\tan{\pi\over18}={\zeta-\zeta^{-1}\over i(\zeta+\zeta^{-1})}\tag7$$ Similarly, 20, 30, 40 degrees are $2\pi/18,3\pi/18,4\pi/18$ radians, respectively, and we get $$\tan{2\pi\over18}={\zeta^2-\zeta^{-2}\over i(\zeta^2+\zeta^{-2})},\quad\tan{3\pi\over18}={\zeta^3-\zeta^{-3}\over i(\zeta^3+\zeta^{-3})},\quad\tan{4\pi\over18}={\zeta^4-\zeta^{-4}\over i(\zeta^4+\zeta^{-4})}\tag8$$ So now you can take the equation in the title, and write it completely in terms of these powers of $\zeta$, and when you multiply through by all the denominators and an appropriate power of $\zeta$, you just get an equation in powers of $\zeta$ that needs to be verified.

At that point, you may need some properties of $\zeta$. By (1), $\zeta^{36}=1$, and $\zeta^{18}=-1$. You may need some more properties of $\zeta$, but we'd have to see the equation first.

And, as suggested in the first line, and verified by Rick's solution, this is the hard way, for this problem. But it's still good to know these methods, as there are situations where they provide the easiest way.

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+1 for elegance. – Rick Decker Jul 18 '12 at 1:02
But unfortunately not clear for me . – Frank Jul 18 '12 at 1:23
What's not clear? What happens when you use the formula just to write the tangent of 10 degrees? Do you know about radian measure? Do you know about complex numbers? Do you know $e^{ix}=\cos x+i\sin x$? I can't help you if you don't give me a clearer picture of what you can and cannot do: "not clear for me" doesn't give me much to go on. – Gerry Myerson Jul 18 '12 at 1:34
@Gerry I suspect that the answer to your last three questions is "no". You and I (and lots of other readers) can appreciate your elegant answer, but it's likely out of the scope of many other readers' background. – Rick Decker Jul 18 '12 at 1:41
Ok Gerry Myerson sorry but my question is give you a picture what exactly i am asking about ,i can't say anything about your solution since i am unable to understand it ,i can't pretend that and say "elegant". – Frank Jul 18 '12 at 1:51

## Another approach:

Lets, start by arranging the expression: $$\tan(20°) \tan(30°) \tan(40°) = \tan(30°) \tan(40°) \tan(20°)$$$$=\tan(30°) \tan(30°+10°) \tan(30° - 10°)$$

Now, we will express $\tan(30° + 10°)$ and $\tan(30° - 10°)$ as the ratio of Prosthaphaeresis Formulas, giving us: $$\tan(30°) \left( \frac{\tan(30°) + \tan(10°)}{1 - \tan(30°) \tan(10°)}\right) \left( \frac{\tan(30°) - \tan(10°)}{1 + \tan(30°) \tan(10°)}\right)$$

$$= \tan(30°) \left( \frac{\tan^2(30°) - \tan^2(10°)}{1 - \tan^2(30°) \tan^2(10°)}\right)$$

Substituting the value of $\color{blue}{\tan(30°)}$,

$$= \tan(30°) \left(\frac{1 - 3\tan²(10°)}{3 - \tan²(10°)}\right)$$

Multiplying and dividing by $\color{blue}{\tan(10°)}$,

$$=\tan(30°) \tan(10°) \left(\frac{1 - 3\tan²(10°)}{3 \tan(10°) - \tan^3(10°)}\right)$$

It can be easily shown that: $\color{blue}{\tan 3A =\large \frac{3 \tan A - \tan^3A}{1-3\tan^2A}}$,

Thus, our problem reduces to: $$=\tan(30°) \tan(10°) \frac{1}{\tan(3\times 10°)}= \tan(10°)$$

QED!

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