Multiple and sub- multiple Angles

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Question

${\left(\frac{\mathrm{sin}{33}^{\circ }}{\mathrm{sin}{11}^{\circ }\mathrm{sin}{49}^{\circ }\mathrm{sin}{71}^{\circ }}\right)}^{2}+{\left(\frac{\mathrm{cos}{33}^{\circ }}{\mathrm{cos}{11}^{\circ }\mathrm{cos}{49}^{\circ }\mathrm{cos}{71}^{\circ }}\right)}^{2}$ is equal to _______.

Moderate
Solution

Given expression is ${\left(\frac{\mathrm{sin}{33}^{\circ }}{\mathrm{sin}{11}^{\circ }\mathrm{sin}\left({60}^{\circ }-{11}^{\circ }\right)\mathrm{sin}\left(60+{11}^{\circ }\right)}\right)}^{2}+{\left(\frac{\mathrm{cos}{33}^{\circ }}{\mathrm{cos}{11}^{\circ }\mathrm{cos}\left({60}^{\circ }-{11}^{\circ }\right)\mathrm{cos}\left({60}^{\circ }+{11}^{\circ }\right)}\right)}^{2}$

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Let $\mathrm{P}\left(\mathrm{k}\right)=\left(1+\mathrm{cos}\frac{\mathrm{\pi }}{4\mathrm{k}}\right)\left(1+\mathrm{cos}\frac{\left(2\mathrm{k}-1\right)\mathrm{\pi }}{4\mathrm{k}}\right)$ $\left(1+\mathrm{cos}\frac{\left(2\mathrm{k}+1\right)\mathrm{\pi }}{4\mathrm{k}}\right)\left(1+\mathrm{cos}\frac{\left(4\mathrm{k}-1\right)\mathrm{\pi }}{4\mathrm{k}}\right)$. Then