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Question Number 26999 by abdo imad last updated on 01/Jan/18

calculate Π_(k=1) ^n cos((a/2^k )) and0<a<π then find the value of lim_(n−>∝) Σ_(k=1) ^n ln(cos((a/2^k ))).

$${calculate}\:\prod_{{k}=\mathrm{1}} ^{{n}} {cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\:\:{and}\mathrm{0}<{a}<\pi\:\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$ $${lim}_{{n}−>\propto} \:\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left({cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\right). \\ $$

Answered by prakash jain last updated on 01/Jan/18

Π_(k=1) ^n cos((a/2^k )) =(1/(sin ((a/2^n ))))sin ((a/2^n ))Π_(k=1) ^n cos((a/2^k )) =((sin a)/(2^n sin ((a/2^n )))) lim_(n→∞) ((sin a)/(2^n sin ((a/2^n )))) lim_(n→∞) ((sin a)/(((sin ((a/2^n )))/(a/2^7 ))∙a))=((sin a)/a)

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\: \\ $$ $$=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}^{{n}} }\right)}\mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}^{{n}} }\right)\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\mathrm{cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\: \\ $$ $$=\frac{\mathrm{sin}\:{a}}{\mathrm{2}^{{n}} \mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}^{{n}} }\right)} \\ $$ $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{sin}\:{a}}{\mathrm{2}^{{n}} \mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}^{{n}} }\right)} \\ $$ $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{sin}\:{a}}{\frac{\mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}^{{n}} }\right)}{\frac{{a}}{\mathrm{2}^{\mathrm{7}} }}\centerdot{a}}=\frac{\mathrm{sin}\:{a}}{{a}} \\ $$ $$ \\ $$

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