2

Find the value of $\lambda$ in $$\dfrac{3-\tan^2 {\pi\over 7}}{1-\tan^2 {\pi\over 7}}=\lambda\cos{\pi\over 7}$$

The numerator looks similar to expansion of $\tan 3x$, so I tried this

$$\dfrac{3\tan {\pi\over 7}-\tan^3 {\pi\over 7}}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$

$$\dfrac{\left(3\tan {\pi\over 7}-\tan^3 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}{\tan {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)\left(1-3\tan^2 {\pi\over 7}\right)}=\lambda\cos{\pi\over 7}$$

$$\dfrac{\tan {3\pi\over 7}\left(1-3\tan^2 {\pi\over 7}\right)}{\sin {\pi\over 7}\left(1-\tan^2 {\pi\over 7}\right)}=\lambda$$

But I'm stuck here. Need help. Thanks in advance.

nonuser
  • 91,557

2 Answers2

6

Let $a= \pi/7$ then we have $$1+{2\over 1-\tan^2a} = \lambda \cos a$$

multiplying this with $\tan a$ we get: $$\tan a+ \tan 2a = \lambda \sin a$$

so $$ \sin a\cos 2a +\sin 2a\cos a = \lambda \sin a \cos a \cos 2a$$

multiplying this with 4 we get $$ 4\sin 3a = \lambda \sin 4a$$

Sinnce $\sin 3a = \sin 4a$ we get $\lambda =4$.

nonuser
  • 91,557
1

Let $7x=\pi$

$$\lambda\cos x=\dfrac{3-\tan^2x}{1-\tan^2x}=\dfrac{3\cos^2x-(1-\cos^2x)}{\cos^2x-(1-\cos^2x)}$$

$$\iff2\lambda\cos^3x-4\cos^2x-\lambda\cos x+1=0$$

From How Can One Prove $\cos(\pi/7) + \cos(3 \pi/7) + \cos(5 \pi/7) = 1/2$,

$\cos\dfrac{(2n+1)\pi}7; n=0,1,2$ are the roots of $8t^3-4t^2-4t+1=0$

Comparing with the coefficients, $$\dfrac{2\lambda}8=\dfrac44=\dfrac\lambda4=\dfrac11$$

$\implies\lambda=?$

So, we can replace $\dfrac\pi7$ with $\dfrac{3\pi}7,\dfrac{5\pi}7$ in the given expression

lab bhattacharjee
  • 279,016