Find the Value of (1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)

Question:

\[ (1+\cos \frac{\pi}{8}) (1+\cos \frac{3\pi}{8}) (1+\cos \frac{5\pi}{8}) (1+\cos \frac{7\pi}{8}) \]

Solution

Use the identity

\[ 1+\cos\theta = 2\cos^2\frac{\theta}{2} \]

Therefore,

\[ P=(1+\cos \frac{\pi}{8}) (1+\cos \frac{3\pi}{8}) (1+\cos \frac{5\pi}{8}) (1+\cos \frac{7\pi}{8}) \] \[ =16 \cos^2\frac{\pi}{16} \cos^2\frac{3\pi}{16} \cos^2\frac{5\pi}{16} \cos^2\frac{7\pi}{16} \] \[ =16 \left( \cos\frac{\pi}{16} \cos\frac{3\pi}{16} \cos\frac{5\pi}{16} \cos\frac{7\pi}{16} \right)^2 \]

Using the standard identity

\[ \cos\frac{\pi}{16} \cos\frac{3\pi}{16} \cos\frac{5\pi}{16} \cos\frac{7\pi}{16} = \frac{\sqrt{2}}{8} \]

Hence,

\[ P = 16\left(\frac{\sqrt2}{8}\right)^2 \] \[ = 16\cdot\frac{2}{64} \] \[ = \frac12 \]

Answer

\[ \boxed{\frac{1}{2}} \]

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