If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$

Question

If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$

My 1st approach :

$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $

$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.

My IInd approach :

U\sing Euler substitution :

\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)

Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$ and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$

By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$

$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$

Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.

Answer 1

Not using Euler's formula which I don't think the best way for this

Let $\displaystyle a=\tan A,b=\tan2A,c=\tan4A$ where $A+2A+4A=n\pi$ where $7\nmid n$

As $\displaystyle\tan(n\pi-rA)=-\tan rA,\tan6A=-\tan A=-a$ etc.

Using Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,

$\displaystyle a+b+c=abc$

Now using Sum of tangent functions where arguments are in specific arithmetic series,

$\displaystyle\tan7x=\frac{\binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x}{1-\binom72\tan^2x+\binom74\tan^4x-\binom76\tan^6x}$

If $\displaystyle\tan7A=0,7A=m\pi$ where $m$ is any integer

$\displaystyle\implies A=\frac{m\pi}7$ where $0\le m\le6$

So, $\pm a,\pm b,\pm c,\tan0=0$ are the roots of $\displaystyle \binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x=0$ $\displaystyle\iff\tan^7x-21\tan^5x+35\tan^3x-7\tan x=0$

So, $\pm a,\pm b,\pm c$ are the roots of $\displaystyle \tan^6x-21\tan^4x+35\tan^2x-7=0\ \ \ \ (1)$

Now the equation whose roots are $\pm a,\pm b,\pm c$ is $\displaystyle(y-a)(y-b)(y-c)(y+a)(y+b)(y+c)=0$ $\displaystyle\iff(y^2-a^2)(y^2-b^2)(y^2-c^2)=0$ $\displaystyle\iff y^6-(a^2+b^2+c^2)y^4+(a^2b^2+b^2c^2+c^2a^2)y^2-a^2b^2c^2=0\ \ \ \ (2)$

If we write $\displaystyle a+b+c=abc=S$ and $\displaystyle ab+bc+ca=T,$

$\displaystyle a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=S^2-2T$

and $\displaystyle a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=T^2-2S^2$

So, $(2)$ becomes $\displaystyle y^6-(S^2-2T)y^4+(T^2-2S^2)y^2-S^2=0$

Comparing with $\displaystyle(1), S^2=7,S^2-2T=21, T^2-2S^2=35$

Can you find the required $T$ from here?

Answer 2

(1) Note first that $$\tan x\tan(2x)=\frac{\sin x\sin(2x)}{\cos x\cos(2x)}=\frac{2\sin^2x}{\cos(2x)}=\frac{1}{\cos(2x)}-1 $$ (2) It follows that $$\eqalign{ S~&\buildrel{\rm def}\over{=}~\tan\alpha\tan{2\alpha}+\tan2\alpha\tan{4\alpha}+\tan4\alpha\tan{\alpha}\cr &=-3+\frac{1}{\cos\alpha}+\frac{1}{\cos2\alpha}+\frac{1}{\cos4\alpha}\cr &=-3+\frac{\cos\alpha\cos2\alpha+\cos2\alpha\cos4\alpha+\cos4\alpha\cos\alpha}{\cos\alpha\cos2\alpha\cos4\alpha}\cr &=-3+\frac{\cos\alpha+\cos3\alpha+\cos6\alpha+\cos2\alpha+\cos5\alpha+\cos 4\alpha}{2\cos\alpha\cos2\alpha\cos4\alpha}\tag{1}\cr } $$ (3) If $\xi=e^{i\alpha}$ then we have $$1+\xi+\xi^2+\xi^3+\xi^4+\xi^5+\xi^6=0$$ Taking real parts we get $$1+\cos\alpha+\cos2\alpha+\cos3\alpha+\cos4\alpha+\cos5\alpha+\cos6\alpha=0\tag{2}$$ also $$8\cos\alpha\cos2\alpha\cos4\alpha=\frac{8\sin\alpha\cos\alpha\cos2\alpha\cos4\alpha}{\sin\alpha}=\frac{\sin8\alpha}{\sin\alpha}=1\tag{3}$$ (4) Replacing $(2)$ and $(3)$ in $(1)$ we obtain $$ S=-3+\frac{-4}{1}=-7. $$ which is the desired conclusion.

Answer 3

$$ sin(a)sin(b)cos(c) = \left(\frac{e^{ja}-e^{-ja}}{2j}\right)\left(\frac{e^{jb}-e^{-jb}}{2j}\right)\left(\frac{e^{jc}+e^{-jc}}{2j}\right) $$ $$ = \left(e^{j(a+b)}+e^{-j(a+b)}-\left(e^{j(a-b)}+e^{-j(a-b)} \right)\right) \left(\frac{e^{jc}+e^{-jc}}{2j}\right) $$ $$ = \left(e^{j(a+b+c)}+e^{-j(a+b+c)}-\left(e^{j(a-b)}+e^{-j(a-b)} \right)\right) \left(\frac{e^{jc}+e^{-jc}}{2j}\right) $$ $$ =\left(e^{j(a+b+c)}+e^{-j(a+b+c)}+e^{j(a+b-c)}+e^{-j(a+b-c)} \right) - \left(e^{j(a-b+c)}+e^{-j(a-b+c)}+e^{j(a-b-c)}+e^{-j(a-b-c)} \right) $$

Now, put $a=\theta, b=2\theta, c=4\theta$, then $a=2\theta, b=4\theta, c=\theta$ and finally $a=4\theta, b=\theta, c=2\theta$. We can see that the terms generated by $(a+b-c)$ and $(a-b+c)$ will cancel out after the substitution. And finally we get, $$ e^{j7\theta}+e^{-j7\theta}+e^{j7\theta}+e^{-j7\theta}+e^{j7\theta}+e^{-j7\theta} - \left( e^{j5\theta}+e^{-j5\theta}+e^{j3\theta}+e^{-j3\theta}+e^{j1\theta}+e^{-j1\theta} \right) $$ $$ = 6 - \left( e^{j5\theta}+e^{-j5\theta}+e^{j3\theta}+e^{-j3\theta}+e^{j1\theta}+e^{-j1\theta} +1 \right) + 1 $$

Using $7\theta = 2\pi$, and $-5\theta = 2\pi-5\theta = 2\theta$, etc., above expression can be rewritten:

$$ \left( e^{j5\theta}+e^{-j5\theta}+e^{j3\theta}+e^{-j3\theta}+e^{j1\theta}+e^{-j1\theta} +1 \right) = \left( e^{j5\theta}+e^{j2\theta}+e^{j3\theta}+e^{j4\theta}+e^{j1\theta}+e^{j6\theta} + e^{j0\theta} \right) = \frac{1-e^{-7j\theta}}{1-e^{-j\theta}} = 0 $$ So, numerator = 7. Similarly, denominator becomes: $$cos(a)cos(b)cos(c) = \left(\frac{e^{ja}+e^{-ja}}{2}\right)\left(\frac{e^{jb}+e^{-jb}}{2}\right)\left(\frac{e^{jc}+e^{-jc}}{2}\right) $$ $$ = \left(e^{j(a+b+c)}+e^{-j(a+b+c)}+e^{j(a+b-c)}+e^{-j(a+b-c)} \right) + \left(e^{j(a-b+c)}+e^{-j(a-b+c)}+e^{j(a-b-c)}+e^{-j(a-b-c)} \right) $$ $$ = \left(e^{j(7\theta)}+e^{-j(7\theta)}+e^{j(-1\theta)}+e^{-j(-1\theta)} \right) + \left(e^{j(1\theta)}+e^{-j(1\theta)}+e^{j(-5\theta)}+e^{-j(-5\theta)} \right) $$ $$ = 2 + \left( e^{j5\theta}+e^{-j5\theta}+e^{j3\theta}+e^{-j3\theta}+e^{j1\theta}+e^{-j1\theta} \right) + 1 - 1 = 1 $$ So tan(a)tan(2a)+tan(2a)tan(4a)+tan(4a)tan(a) = -7 (minus comes because while the 2's in the denominators of sin and cos Euler formulas cancel, the $j^2$ term from sin remains, resulting in negative sign.

Answer 4

We can usually solve these kind of problems by building an equation for which the the given trigonometric values are the roots. (See S.L.Loney's book)

It's as much time consuming as doing it by expanding and simplifying the angle sums, but we get several other answer's easily once the equation is in place.

Here is an overview of the method:

First, use (can be derived from Euler's formula):

\begin{align*} \cos{(7\, \alpha)} &= 64\cos(\alpha)^7-112\cos(\alpha)^5+56\cos(\alpha)^3-7\cos(\alpha) \end{align*}

Substituting for $\alpha$ and factoring gives:

\begin{align*} 8c^3+4c^2-4c-1 &= 0 \tag 1 \end{align*}

which is the equation whose roots are $\cos\left(\frac{2\pi}{7}\right),\cos\left(\frac{4\pi}{7}\right)$ and $\cos\left(\frac{2\pi}{7}\right)$

Substituting $c=\frac{1}{c}$ in $(1)$ gives:

\begin{align*} c^3+4c^2-4c-8 &= 0 \tag 2 \end{align*}

which has the roots $\sec \left(\frac{2\pi}{7}\right),\sec\left(\frac{4\pi}{7}\right)$ and $\sec\left(\frac{2\pi}{7}\right)$

Substituting $c=\sqrt{1+c}$ in $(2)$ and rationalizing gives:

\begin{align*} c^3-21c^2+35c-7 &= 0 \tag 3 \end{align*}

which has the roots $\tan \left(\frac{2\pi}{7}\right)^2 ,\tan\left(\frac{4\pi}{7}\right)^2$ and $\tan\left(\frac{2\pi}{7}\right)^2$

Getting further equations where roots will be square of the current roots is easy, but going back to find square roots is a bit ambiguous for me, I'd like to know if there is a better way. I did it like this:

Consider an equation, \begin{align*} x^3-a\, x^2+b\, x - c &= 0 \tag 4 \end{align*}

Substitute $x=\sqrt{x}$ and rationalize:

\begin{align*} x^3-(a^2 - 2\, b)x^2+(b^2 - 2ac)\, x - c^2 &= 0 \tag 5 \end{align*}

Compare with the equation $(2)$ (subst. $c=x$ there)

\begin{align*} x^3-21x^2+35x-7 &= 0 \tag 6 \end{align*}

and solve for $a,b,c$ and choose by inspection:

$(4)$ would be: \begin{align*} x^3+\sqrt{7}x^2-7x+\sqrt{7} &= 0 \tag 7 \end{align*}

which is the equation having $\tan \left(\frac{2\pi}{7}\right) ,\tan\left(\frac{4\pi}{7}\right)$ and $\tan\left(\frac{2\pi}{7}\right)$ as the roots.

Hence, the required sum:

\begin{align*} \tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha &= -7 \end{align*}

From this equation, we can obtain several other identites, such as:

\begin{align*} \tan\alpha +\tan2\alpha + \tan4\alpha &= -\sqrt{7} \\ \tan\alpha \cdot \tan2\alpha \cdot \tan4\alpha &= -\sqrt{7} \\ \end{align*}

And from $(6)$,

\begin{align*} \\ \tan\left(\alpha\right)^2 +\tan\left(2\alpha\right)^2 + \tan\left(4\alpha\right)^2 &= 21 \\ \left(\tan\alpha \tan2\alpha\right)^2 +\left(\tan2\alpha \tan4\alpha\right)^2 +\left(\tan4\alpha \tan\alpha\right)^2 &= 35 \\ \tan\left(\alpha\right)^2 \cdot \tan\left(2\alpha\right)^2 \cdot \tan\left(4\alpha\right)^2 &= 7 \\ \end{align*}

and so on for powers of $4, 8$ etc.