Trigonometry question

Question

If $$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}.$$ If $\alpha$ is a natural number.Find $\alpha$.

My attempt is:

$$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$

convert it into sin,cos

$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$

$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos\frac{2\pi}{7}}=\alpha \cos\frac{\pi}{7}$$

$$3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$

$$2\cos^2\frac{\pi}{7}+\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$

$$2\cos^2\frac{\pi}{7}+\cos\frac{2\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$

but i got stuck and could not further solve it.... I would appreciate the help,thanks in advance.

see here http://www.wolframalpha.com/input/?i=%283-tan%28pi%2F7%29%5E2%29%2F%281%2Btan%28pi%2F7%29%5E2%29%2B%281%2Fcos%28pi%2F7%29%29 — Dr. Sonnhard Graubner, Jul 25 '15 at 12:56
see also here http://mathworld.wolfram.com/TrigonometryAnglesPi7.html — Dr. Sonnhard Graubner, Jul 25 '15 at 13:00

score 5 · Accepted Answer · answered Jul 25 '15 at 13:13

5

$$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=1+\frac{2}{1-\tan^2\frac{\pi}7}=1+\frac{2\tan\frac{\pi}7}{1-\tan^2\frac{\pi}7}\frac1{\tan\frac{\pi}7}=1+\frac{\tan\frac{2\pi}{7}}{\tan\frac{\pi}7}\\=1+\frac{\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=\frac{\sin\frac{3\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=\frac{\sin\frac{4\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=4\cos\frac{\pi}7.$$

answered Jul 25 '15 at 13:13

Eclipse Sun

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Thank you very much @Eclipse Sun. – Vinod Kumar Punia Jul 25 '15 at 13:17
Is $4\cos \frac \pi 7$ natural number ??! – Mohamed Mostafa Jul 25 '15 at 14:17
@Mohamed_Mostafa look into the question and see that $\alpha$ is multiply with $\cos \frac{\pi}{7}$ – IrbidMath Jul 25 '15 at 15:05

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

Use $\tan^2A=\cdots=\dfrac{1-\cos^2A}{\cos^2A}$ and replace $\dfrac\pi7$ with $B$

to get $$2\alpha\cos^3B-4\cos^2B-\alpha\cos B+1=0\ \ \ \ (1)$$

Now if $7C=(2n+1)\pi$ where $n$ is any integer

$C=\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$

and $\cos4C=\cdots=-\cos3C\implies8\cos^4C-8\cos^2C+1=-(4\cos^3C-3\cos C)$

$\iff8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1=0\ \ \ \ (2)$

whose roots will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$

Now $n\equiv3\implies\cos\dfrac{(2n+1)\pi}7=\cdots=-1$

So, the roots of $0=\dfrac{8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1}{\cos C+1}=8\cos^3C-4\cos^2C-4\cos A+1\ \ \ \ (3)$

will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$

Now compare $(1),(3)$ to get $\dfrac{2\alpha}8=\dfrac44=\dfrac{\alpha}4=\dfrac11$

Observation: The given problem holds true for $\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$

$(3)$ can be derived like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$

Trigonometry question

2 Answers2