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I need help solving this question: $$ \text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.} $$ I simplified this down to:

$$ \frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1} $$

But am unable to proceed further. The value of k is given to be 4, but I am unable to derive that result.

Kindly provide me with some insight, or with a step-by-step solution.

Thanks in advance,

Abhigyan

Abhigyan
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4 Answers4

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\begin{align} k&=\frac{4\cos^2\frac\pi7-1}{\cos\frac{\pi}{7}(2\cos^2\frac\pi7-1)}\\ &=\frac{2\cos\frac{2\pi}{7}+1}{\cos\frac{2\pi}{7}\cos\frac{\pi}{7}}\\ &=\frac{2\cos\frac{2\pi}{7}+1}{\frac{1}{2}(\cos\frac{3\pi}{7}+\cos\frac{\pi}{7})}\\ &=\frac{4\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{3\pi}{7}+\sin\frac{\pi}{7}\cos\frac{\pi}{7}}\\ &=\frac{2\sin\frac{3\pi}{7}-2\sin\frac{\pi}{7}+2\sin\frac{\pi}{7}}{\frac{1}{2}(\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}+\sin\frac{2\pi}{7})}\\ &=\frac{4\sin\frac{3\pi}{7}}{\sin\frac{4\pi}{7}}\\ &=4 \end{align}

CY Aries
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We need to prove that $$3-\tan^2\frac{\pi}{7}=4(1-\tan^2\frac{\pi}{7})\cos\frac{\pi}{7}$$ or $$\frac{3}{2}+\frac{3}{2}\cos\frac{2\pi}{7}-\frac{1}{2}+\frac{1}{2}\cos\frac{2\pi}{7}=4\cos\frac{2\pi}{7}\cos\frac{\pi}{7}$$ or

$$1+\frac{3}{2}\cos\frac{2\pi}{7}+\frac{1}{2}\cos\frac{2\pi}{7}=2\cos\frac{3\pi}{7}+2\cos\frac{\pi}{7}$$ or $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$ or $$2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}=-2\sin\frac{\pi}{7}$$ or $$\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}=-\sin\frac{\pi}{7}.$$ Id est, indeed, $k=4$.

Done!

Michael Rozenberg
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$$k\cos y=\dfrac{3-\tan^2y}{1-\tan^2y}=\dfrac{3\cos^2y-\sin^2y}{\cos^2y-\sin^2y}=\dfrac{4\cos^2y-1}{2\cos^2y-1}$$

$$\iff2k\cos^3y-4\cos^2y-k\cos y+1=0\ \ \ \ (1)$$

Now if $7y=(2n+1)\pi,$

$\cos4y=\cdots=-\cos3y$

Using $\cos2A=2\cos^2A-1,\cos3B=4\cos^3B-3\cos B$

The roots of $$8\cos^4y+4\cos^3y-8\cos^2y-3\cos y+1=0$$ are $$\cos\dfrac{(2n+1)\pi}7$$ where $n\equiv0,\pm1,\pm2,\pm3\pmod7$

As for $n=3,\cos\dfrac{(2n+1)\pi}7=-1,$ the roots of $$0=\dfrac{8\cos^4y+4\cos^3y-8\cos^2y-3\cos y+1}{\cos y+1}$$ $$\iff8\cos^3y-4\cos^2y-4\cos y+1=0\ \ \ \ (2)$$ will be $$\cos\dfrac{(2n+1)\pi}7$$ where $n\equiv0,\pm1,\pm2,3\pmod7$

Compare $(1),(2)$

lab bhattacharjee
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Set $z=e^{i\pi/7}$, so $z^7=e^{i\pi}=-1$ and $z^{-1}=\bar{z}$.

Then $$ \cos\frac{\pi}{7}=\frac{z+\bar{z}}{2}=\frac{z^2+1}{2z}, \qquad \sin\frac{\pi}{7}=\frac{z-\bar{z}}{2i}=\frac{z^2-1}{2iz} $$ Therefore $$ \tan\frac{\pi}{7}=-i\frac{z^2-1}{z^2+1} $$ Plugging in the left hand side, we get $$ \frac{3(z^2+1)^2+(z^2-1)^2}{(z^2+1)^2+(z^2-1)^2}= \frac{4z^4+4z^2+4}{2z^4+2}=2\frac{z^4+z^2+1}{z^4+1} $$ and finally $$ k=2\frac{z^4+z^2+1}{z^4+1}\frac{2z}{z^2+1}= \frac{4z(z^4+z^2+1)}{(z^4+1)(z^2+1)} $$ The denominator reads $z^6+z^4+z^2+1$, but from $z^7+1=0$ we can deduce $z^6-z^5+z^4-z^3+z^2-z+1=0$ and so $$ z^6+z^4+z^2+1=z^5+z^3+z=z(z^4+z^2+1) $$ and you get $$ k=4 $$

egreg
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