Question

Using the identity, $sin^4x = \frac{3}{8} - \frac{1}{2}cos2x + \frac{1}{8}cos4x$ or otherwise, if the value of $sin^4(\frac{\pi}{7}) + sin^4(\frac{3\pi}{7}) + sin^4(\frac{5\pi}{7}) = \frac{a}{b}$ where a and b are coprime, then the value of (a - b) is

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (D)

5

Answer 2

The sum is $S = sin^4(\frac{\pi}{7}) + sin^4(\frac{3\pi}{7}) + sin^4(\frac{5\pi}{7})$. Using the identity $sin^4x = \frac{3}{8} - \frac{1}{2}cos(2x) + \frac{1}{8}cos(4x)$: $S = \sum_{x \in \{\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}\}} (\frac{3}{8} - \frac{1}{2}cos(2x) + \frac{1}{8}cos(4x))$ $S = 3 \times \frac{3}{8} - \frac{1}{2} \sum cos(2x) + \frac{1}{8} \sum cos(4x)$ $S = \frac{9}{8} - \frac{1}{2} (cos(\frac{2\pi}{7}) + cos(\frac{6\pi}{7}) + cos(\frac{10\pi}{7})) + \frac{1}{8} (cos(\frac{4\pi}{7}) + cos(\frac{12\pi}{7}) + cos(\frac{20\pi}{7}))$ Using the property that for $n=7$, $\sum_{k=1}^{n-1} cos(\frac{2k\pi}{n}) = -1$ and $\sum_{k=1}^{n-1} cos(\frac{4k\pi}{n}) = -1$. The sums of cosines are: $cos(\frac{2\pi}{7}) + cos(\frac{6\pi}{7}) + cos(\frac{10\pi}{7}) = cos(\frac{2\pi}{7}) + cos(\frac{6\pi}{7}) + cos(\frac{4\pi}{7}) = -\frac{1}{2}$ $cos(\frac{4\pi}{7}) + cos(\frac{12\pi}{7}) + cos(\frac{20\pi}{7}) = cos(\frac{4\pi}{7}) + cos(\frac{2\pi}{7}) + cos(\frac{6\pi}{7}) = -\frac{1}{2}$ $S = \frac{9}{8} - \frac{1}{2}(-\frac{1}{2}) + \frac{1}{8}(-\frac{1}{2}) = \frac{9}{8} + \frac{1}{4} - \frac{1}{16} = \frac{18+4-1}{16} = \frac{21}{16}$. Given $S = \frac{a}{b}$, we have $a=21$ and $b=16$. $a-b = 21-16 = 5$.