Trigonometry
Product of cosine half-angle expressions
MJAT_TS6_P1
Grade 12
Question:
The value of $8\left(\frac{1}{2}+\cos\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos\frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos\frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos\frac{27\pi}{20}\right)$ is equal to:
Step-by-Step Solution
Key Concept: Use the identity $\frac{1}{2}+\cos\theta = \frac{1+2\cos\theta}{2}$ and the Chebyshev product formula for $\frac{\sin(n\theta)}{\sin\theta}=2^{n-1}\prod\cos(k\theta)$.
Value $=\mathbf{0.50}$.
Correct Answer: 0.50