Evaluate [1 + cos (π/8)][1 + cos (3π/8)][1 + cos (5π/8)][1 + cos (7π/8)]

Rumus terpenting untuk menyelesaikan ini adalah rumus rasio relasi trigonometri cos [(π/2) – θ] = sin θ.

\displaystyle \left ( 1+\cos \frac{\pi}{8} \right )\left ( 1+\cos \frac{3\pi}{8} \right )\left ( 1+\cos \frac{5\pi}{8} \right )\left ( 1+\cos \frac{7\pi}{8} \right )\\ =\left ( 1+\cos \frac{\pi}{8} \right )\left ( 1+\cos \frac{3\pi}{8} \right )\left [ 1+\cos \left ( \pi-\frac{3\pi}{8} \right ) \right ]\left [ 1+\cos \left ( \pi-\frac{\pi}{8} \right ) \right ]\\ =\left ( 1+\cos \frac{\pi}{8} \right )\left ( 1+\cos \frac{3\pi}{8} \right )\left ( 1-\cos \frac{3\pi}{8} \right )\left ( 1-\cos \frac{\pi}{8} \right )\\=\left ( 1-\cos^2 \frac{\pi}{8} \right )\left ( 1-\cos^2 \frac{3\pi}{8} \right )\\ =\sin^2 \frac{\pi}{8}\left [ 1-\cos^2 \left ( \frac{\pi}{2}-\frac{\pi}{8} \right ) \right ]\\ =\sin^2 \frac{\pi}{8}\left ( 1-\sin^2 \frac{\pi}{8} \right )\\ =\sin^2 \frac{\pi}{8}~\cos^2 \frac{\pi}{8}\\=\frac{1-\cos \frac{\pi}{4}}{2}~\frac{1+\cos \frac{\pi}{4}}{2}\\ =\frac{1-\cos^2 \frac{\pi}{4}}{4}\\ =\frac{1-\left ( \frac{\sqrt{2}}{2} \right )^2}{4}\\=\frac{1}{8}\\ =0,125
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