Question

$4 \sin \frac{2\pi}{7} - \tan \frac{\pi}{7} = \sqrt{7}$.

Answer 1

True

Answer 2

To verify the identity $4 \sin \frac{2\pi}{7} - \tan \frac{\pi}{7} = \sqrt{7}$, we will use a known trigonometric identity related to angles of the form $\frac{k\pi}{n}$.

Key Identities for $n=7$:

1. The product of sines: $\prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^{n-1}}$. For $n=7$: $\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} \sin \frac{4\pi}{7} \sin \frac{5\pi}{7} \sin \frac{6\pi}{7} = \frac{7}{2^{7-1}} = \frac{7}{64}$. Since $\sin(\pi - x) = \sin x$, we have: $\sin \frac{4\pi}{7} = \sin (\pi - \frac{3\pi}{7}) = \sin \frac{3\pi}{7}$ $\sin \frac{5\pi}{7} = \sin (\pi - \frac{2\pi}{7}) = \sin \frac{2\pi}{7}$ $\sin \frac{6\pi}{7} = \sin (\pi - \frac{\pi}{7}) = \sin \frac{\pi}{7}$ So, the product becomes: $(\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7})^2 = \frac{7}{64}$ Taking the square root (all sines are positive for angles in $(0, \pi)$): $\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} = \frac{\sqrt{7}}{8}$. (Equation 1)

2. The product of cosines: $\prod_{k=1}^{n-1} \cos \frac{k\pi}{n} = \frac{\sin(n\pi/2)}{2^{n-1}}$ for odd $n$. For $n=7$: $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7} \cos \frac{4\pi}{7} \cos \frac{5\pi}{7} \cos \frac{6\pi}{7} = \frac{\sin(7\pi/2)}{2^{7-1}} = \frac{\sin(3\pi + \pi/2)}{64} = \frac{-\sin(\pi/2)}{64} = -\frac{1}{64}$. Since $\cos(\pi - x) = -\cos x$, we have: $\cos \frac{4\pi}{7} = \cos (\pi - \frac{3\pi}{7}) = -\cos \frac{3\pi}{7}$ $\cos \frac{5\pi}{7} = \cos (\pi - \frac{2\pi}{7}) = -\cos \frac{2\pi}{7}$ $\cos \frac{6\pi}{7} = \cos (\pi - \frac{\pi}{7}) = -\cos \frac{\pi}{7}$ So, the product becomes: $(\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}) (-\cos \frac{3\pi}{7}) (-\cos \frac{2\pi}{7}) (-\cos \frac{\pi}{7}) = -\frac{1}{64}$ $-(\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7})^2 = -\frac{1}{64}$ $(\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7})^2 = \frac{1}{64}$ Taking the square root (all cosines are positive for angles in $(0, \pi/2)$, here $\frac{\pi}{7}, \frac{2\pi}{7}, \frac{3\pi}{7}$ are all in $(0, \pi/2)$): $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7} = \frac{1}{8}$. (Equation 2)

Deriving the Tangent Product Identity: Divide Equation 1 by Equation 2: $\frac{\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7}}{\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}} = \frac{\sqrt{7}/8}{1/8}$ $\tan \frac{\pi}{7} \tan \frac{2\pi}{7} \tan \frac{3\pi}{7} = \sqrt{7}$. (Equation 3)

Verifying the Given Identity: Let $A = \frac{\pi}{7}$. The identity to verify is $4 \sin 2A - \tan A = \sqrt{7}$. Substitute $\sqrt{7}$ from Equation 3 into the identity: $4 \sin 2A - \tan A = \tan A \tan 2A \tan 3A$ Rearrange the terms: $4 \sin 2A = \tan A + \tan A \tan 2A \tan 3A$ $4 \sin 2A = \tan A (1 + \tan 2A \tan 3A)$ Express tangents in terms of sines and cosines: $4 \sin 2A = \frac{\sin A}{\cos A} \left(1 + \frac{\sin 2A \sin 3A}{\cos 2A \cos 3A}\right)$ $4 \sin 2A = \frac{\sin A}{\cos A} \left(\frac{\cos 2A \cos 3A + \sin 2A \sin 3A}{\cos 2A \cos 3A}\right)$ Use the cosine difference formula: $\cos X \cos Y + \sin X \sin Y = \cos(X-Y)$. $4 \sin 2A = \frac{\sin A}{\cos A} \frac{\cos(3A - 2A)}{\cos 2A \cos 3A}$ $4 \sin 2A = \frac{\sin A}{\cos A} \frac{\cos A}{\cos 2A \cos 3A}$ Cancel $\cos A$: $4 \sin 2A = \frac{\sin A}{\cos 2A \cos 3A}$ Multiply both sides by $\cos 2A \cos 3A$: $4 \sin 2A \cos 2A \cos 3A = \sin A$ Use the double angle formula $2 \sin X \cos X = \sin 2X$: $2 (2 \sin 2A \cos 2A) \cos 3A = \sin A$ $2 \sin 4A \cos 3A = \sin A$ Use the product-to-sum formula $2 \sin X \cos Y = \sin(X+Y) + \sin(X-Y)$: $\sin(4A+3A) + \sin(4A-3A) = \sin A$ $\sin 7A + \sin A = \sin A$ Subtract $\sin A$ from both sides: $\sin 7A = 0$ Since $A = \frac{\pi}{7}$, we have $7A = 7 \times \frac{\pi}{7} = \pi$. $\sin \pi = 0$. This is a true statement. Therefore, the original identity is correct.