4

How do I find the value of the following expression? $$ \left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $$

I tried writing the numerator as $\tan 40^\circ - \tan80^\circ -\tan20^\circ,$ but then the expression was getting complicated.

Blue
  • 75,673
AtharvaSakpal
  • 49
  • Do you have to reduce all angles to between 0 and 90 degrees? If you were using a table of trig functions you might have to do that but using the "Windows" calculator tan(40)+ tan(100)+ tan(160)= 0.83909963117728001176312729812318-5.671281819617709530994418439864-0.36397023426620236135104788277683= -5.1961524227066318805823390245176 – user247327 Aug 19 '20 at 14:09
  • $$\begin{align} & \text{You have:} \ & \frac{\tan40^\circ + \tan(-80^\circ) + \tan(-20^\circ)}{\tan(-20^\circ)\tan40^\circ\tan(-80^\circ)} \ {} \ & \text{Then ultiplying the numerator and denominator} \ & \text{by the product of three cosines yields this:} \ = {} & \frac{\sin40^\circ\cos(-20^\circ)\cos(-80^\circ) + \sin(-80^\circ)\cos40^\circ\cos(-20^\circ) + \sin(-20^\circ)\cos40^\circ\cos(-80^\circ)}{\sin(-20^\circ)\sin40^\circ \sin(-80^\circ)} \end{align}$$ – Michael Hardy Aug 19 '20 at 14:14
  • The numerator is $$\begin{align} & \sin a\cos b\cos c + \cos a\sin b\cos c + \cos a \cos b \sin c \ {} \ = {} & -\Big(\sin a \sin b \sin c -\sin a\cos b\cos c - \cos a\sin b\cos c - \cos a \cos b \sin c\Big) + \sin a\sin b\sin c \[8pt] = {} & -\sin(a+b+c) + \sin a\sin b \sin c \[8pt] = {} & -\sin(-60^\circ) + \sin a \sin b \sin c \end{align} $$ – Michael Hardy Aug 19 '20 at 14:15
  • Then let $A:=\sin(20^\circ)\sin(40^\circ)\sin(80^\circ)$, consider $A\cos(20^\circ)=\frac{1}{8}\sin(160^\circ)$ thus $A=\frac18\tan(20^\circ)$ so you have the denominator done. – Alexey Burdin Aug 19 '20 at 14:21

3 Answers3

3

First of all we have (see Morrie's law) $$ \tan20^\circ\tan40^\circ\tan80^\circ=\sqrt{3}. $$ The numerator is, setting $x=20^\circ,$ \begin{align} &\tan40^\circ+\tan100^\circ+\tan160^\circ=\\ &\qquad\qquad=\tan(60^\circ-x)+\tan(120^\circ-x)+\tan(180^\circ-x)=\\ &\qquad\qquad= \frac{\tan 60^\circ-\tan x}{1+\tan 60^\circ\tan x}+ \frac{\tan120^\circ-\tan x}{1+\tan120^\circ\tan x}+ \frac{\tan180^\circ-\tan x}{1+\tan180^\circ\tan x}=\\ &\qquad\qquad= \frac{ \sqrt{3}-\tan x}{1+\sqrt{3}\tan x}+ \frac{-\sqrt{3}-\tan x}{1-\sqrt{3}\tan x}- \tan x=\\ &\qquad\qquad= \frac{\sqrt{3}\cos x-\sin x}{\cos x+\sqrt{3}\sin x}- \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}- \frac{\sin x}{\cos x}=\\ &\qquad\qquad= -3\cdot\frac{3\sin x\cos^2 x-\sin^3 x}{\cos^3 x-3\sin^2 x\cos x}=\\ &\qquad\qquad=-3\cdot\frac{\sin(3x)}{\cos(3x)}=-3\tan60^\circ=-3\sqrt{3} \end{align}

So the final result is $$ \left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|=\left|\frac{-3\sqrt{3}}{\sqrt{3}}\right|=3 $$

Vincenzo Tibullo
  • 10,955
1

@enzotib Already precisely answered your question, while I've developed severe craziness trying to solve this problem. Nevertheless, I want to show you my method, which is based on finding the best approximation of this expression. ( Remember, this is based on approximation, so we're not going to get a perfect value ). The idea is to keep calculations simple so you can find the value of this expression without a calculator.

Let's start with the numerator:

$tan40+tan100+tan160 = tan40-tan80-tan20$ because 100-80 and 160-20 are pairs of related angles.

I will leave the denominator as it is.

Now we must find the tangents of these angles. Let's make use of the small angle approximation, according to which $tanx = x$ for small angles measured in radians ( of course the smaller the angle is, the better the approximation is ). I will now apply this approximation to find a value for tan20°, but first I need to convert 20° in radians.

20 is $\frac{180}{9}$ and $180$ is $\pi$ radians, so $20° = \frac{\pi}{9} \approx tan20°$ because of the small angle approximation.

To avoid long and difficult calculations, it is better if we try to find a value for $\frac{\pi}{9}$.

$\pi \approx 3$,so $\frac{\pi}{9} \approx \frac{3}{9} \approx \frac{1}{3} \approx 0.33$.

Keep in mind that we're starting with a rounded down value, so the next time we need an approximation, if the situation allows it, we should use a rounded up value

Let's now find tan40° with the double angle formula:

$tan2x= \frac{2tanx}{1-tan^2x}$

$tan40=\frac{2tan20}{1-tan^220} = \frac{2*0.33}{1-(0,33)^2} = \frac{0.66}{1-0.1} = \frac{0,66}{0,9} \approx \frac{0,7}{0.9} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{9} = 0,77 \approx 0,8$

Now we rounded up two times, when approximating the fraction and when approximating the final value, so next time we will round down if possible

Now I will apply the double angle formula again to find $tan80$

$tan80 = \frac{2tan40}{1-tan^240} = \frac{2*0.8}{1-(0.8)^2} = \frac{1.6}{1-0.64} = \frac{1.6}{0,36} \approx \frac{1,6}{0,4} = 4 $

We approximated 0.36 as 0.4, so we rounded down a bit because the denominator got bigger.

We can now find a value for the original expression:

Numerator:

$tan40-tan80-tan20 = 0,8-4-0,33=-4,33+0,8=-3,53 $

Denominator:

$tan20tan40tan80 = 0,8*4*0,33 = 3,2*0,33 \approx 3,2*0,3 = 3,2*\frac{3}{10} = \frac{9,6}{10} = 0,96 $

Now let's recompose the fraction:

$\frac{3,53}{0,96}$

We're in this situation: We started with a slightly rounded down value for tan20°, then we rounded up tan40° two times ( heavy round-up ) and then rounded down tan80° a little. The numerator is -tan20°+tan40°-tan80° this means that +tan40° should be an excess value because tan40° was approximated by excess two times while tan80° was approximated by defect slightly, then I add tan20° that is a value approximated by defect, so the numerator should be a too-high value.

Now let's analyze the denominator: 0.8 is an excess value ( tan40°) while 4 is a slightly lower value than the real one, so 4*0,8 is a little bit too high, but I multiplied it by a less than zero slightly lower value so we should be a little bit lower + the final value was rounded down so I should have only a very slight excess

Numerator: mid-high excess Denominator: very low excess-balanced

Overall: mid-high excess

A low excess in the denominator means that the fraction is slightly approximated by defect, but now we have to add the mid-high excess in the numerator, so overall this fraction has an higher-than normal value, so if possible, we're gonna try to round down.

( I removed the - sign in the final fraction because the expression has an absolute value )

Final answer by taking account of all these approximations should be:

$\frac{3,53}{0,96} \approx \frac{3,5}{1} \approx 3,5 $

I approximated in a way such that the numerator is slightly lower and the numerator slightly bigger in an attempt to compensate my excessive round-up.

As you can see the value is slightly off, because it should be 3, that's probably because of my aggressive round-up when dealing with tan40°

Yes this is not a precise answer, this is just my attempt at approximating the value of this expression, of course the precise answer posted by @enzotib is much better

Tom Avery
  • 101
1

Notice that each $\newcommand{\degree}{{\lower{.5pt}\Large\circ}}x\in\left\{20^\degree,-40^\degree,80^\degree\right\}$ satisfies $$ \begin{align} \sqrt3 &=\tan(3x)\\ &=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}\tag1 \end{align} $$ Thus, $$ \tan^3(x)-3\sqrt3\tan^2(x)-3\tan(x)+\sqrt3=0\tag2 $$ Vieta says that the sum of the roots is the negative of the coefficient of $\tan^2(x)$. That is, $$ \tan\left(20^\degree\right)-\tan\left(40^\degree\right)+\tan\left(80^\degree\right)=3\sqrt3\tag3 $$ and the product of the roots is the negative of the constant term. That is, $$ -\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)=-\sqrt3\tag4 $$ Therefore, $$ \begin{align} \frac{\tan\left(40^\degree\right)+\tan\left(100^\degree\right)+\tan\left(160^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)} &=\frac{\tan\left(40^\degree\right)-\tan\left(80^\degree\right)-\tan\left(20^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)}\\[6pt] &=-3\tag5 \end{align} $$ Just take the absolute value of $(5)$.

robjohn
  • 345,667