Question

Prove that cosπ/11 cos2π/11 cos3π/11 cos4π/11 cos5π/11=1/32

Answer 1

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Answer 2

To prove: $\cos\left(\frac{\pi}{11}\right)\cdot\cos\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)=\dfrac{1}{32}$

This is trigonometry prove. Taking left side and to prove the right side.

$\Rightarrow \cos\left(\frac{\pi}{11}\right)\cdot\cos\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Multiply by $2\sin\left(\frac{\pi}{11}\right)$ at numerator and denominator

$\Rightarrow \dfrac{1}{2\sin\left(\frac{\pi}{11}\right)}2\sin\left(\frac{\pi}{11}\right)\cdot \cos\left(\frac{\pi}{11}\right)\cdot\cos\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Formula: 2sin a cos a = sin(2a)

$\Rightarrow \dfrac{1}{2\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Multiply by 2 at numerator and denominator

$\Rightarrow \dfrac{1}{2^2\sin\left(\frac{\pi}{11}\right)}2\sin\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{2\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Formula: 2sin a cos a = sin(2a)

$\Rightarrow \dfrac{1}{2^2\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{4\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Multiply by 2 at numerator and denominator and 2sin a cos a = sin(2a)

$\Rightarrow \dfrac{1}{2^3\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{8\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

$\sin\left(\frac{8\pi}{11}\right)=\sin\left(\pi-\frac{3\pi}{11}\right)=\sin\left(\frac{3\pi}{11}\right)$

$\Rightarrow \dfrac{1}{2^3\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{3\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Multiply by 2 at numerator and denominator and 2sin a cos a = sin(2a)

$\Rightarrow \dfrac{1}{2^4\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{6\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

$\sin\left(\frac{6\pi}{11}\right)=\sin\left(\pi-\frac{5\pi}{11}\right)=\sin\left(\frac{5\pi}{11}\right)$

$\Rightarrow \dfrac{1}{2^4\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{5\pi}{11}\right)\cdot\cos\left(\frac{5\pi}{11}\right)$

Multiply by 2 at numerator and denominator and 2sin a cos a = sin(2a)

$\Rightarrow \dfrac{1}{2^5\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{10\pi}{11}\right)$

$\Rightarrow \dfrac{1}{2^5\sin\left(\frac{\pi}{11}\right)}\sin\left(\pi-\frac{\pi}{11}\right)$

$\Rightarrow \dfrac{1}{2^5\sin\left(\frac{\pi}{11}\right)}\sin\left(\frac{\pi}{11}\right)$

$\Rightarrow \dfrac{1}{2^5}$

$\Rightarrow \dfrac{1}{32}$

Hence Proved