Trigonometry & Inverse Trigonometry
Trig Ratios Functions Identities
nta_abhyas_2025
Grade 11

Question:

The value of $(1 + \cos \frac{2\pi}{7})(1 + \cos \frac{4\pi}{7})(1 + \cos \frac{6\pi}{7})$ is equal to
$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{16}$

Step-by-Step Solution

Key Concept: Use supplementary and supplementary angle identities combined with difference of squares to simplify products of cosine terms
The product $(1 + \cos\frac{\pi}{8})(1 + \cos\frac{7\pi}{8})(1 + \cos\frac{9\pi}{8})(1 + \cos\frac{15\pi}{8})$ can be rewritten using the fact that $\cos(\pi - x) = -\cos(x)$ and $\cos(\pi + x) = -\cos(x)$. This gives $(1 + \cos\frac{\pi}{8})(1 - \cos\frac{\pi}{8})(1 - \cos\frac{\pi}{8})(1 + \cos\frac{\pi}{8})$. Using the difference of squares formula $(1 - \cos^2\frac{\pi}{8})^2 = \sin^4\frac{\pi}{8}$. After simplification using half-angle formulas and the identity $\sin\frac{\pi}{8} = \frac{1}{2}\sin\frac{\pi}{4}$, the answer equals $\frac{1}{8}$.
Correct Answer: 1

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