Question Number 72447 by Maclaurin Stickker last updated on 28/Oct/19
$${Find}\:{the}\:{value}\:{of}: \\ $$$$\frac{\mathrm{1}}{{sin}\:\mathrm{2}°}+\frac{\mathrm{1}}{{sin}\:\mathrm{4}°}+\frac{\mathrm{1}}{{sin}\:\mathrm{8}°}+\frac{\mathrm{1}}{{sin}\:\mathrm{16}°}+... \\ $$$$+\frac{\mathrm{1}}{{sin}\:\left(\mathrm{2}^{\mathrm{1029}} \right)^{°} } \\ $$
Answered by Tanmay chaudhury last updated on 28/Oct/19
$${let} \\ $$$$ \\ $$$${a}=\mathrm{2}^{{o}} \:\:\:{and}\:{n}=\mathrm{1029} \\ $$$${S}={coseca}+{cosec}\mathrm{2}{a}+{cosec}\mathrm{2}^{\mathrm{2}} {a}+...+{cosec}\mathrm{2}^{{n}−\mathrm{1}} {a} \\ $$$${now}\:{cota}−{cot}\frac{{a}}{\mathrm{2}} \\ $$$$=\frac{{cosa}}{{sina}}−\frac{{cos}\frac{{a}}{\mathrm{2}}}{{sin}\frac{{a}}{\mathrm{2}}} \\ $$$$=\frac{{sin}\frac{{a}}{\mathrm{2}}{cosa}−{cos}\frac{{a}}{\mathrm{2}}{sina}}{{sina}.{sin}\frac{{a}}{\mathrm{2}}}=\frac{{sin}\left(\frac{{a}}{\mathrm{2}}−{a}\right)}{{sina}.{sin}\frac{{a}}{\mathrm{2}}}=−{coseca} \\ $$$$\boldsymbol{{coseca}}=\boldsymbol{{cot}}\frac{\boldsymbol{{a}}}{\mathrm{2}}−\boldsymbol{{cota}} \\ $$$$\boldsymbol{{cosec}}\mathrm{2}\boldsymbol{{a}}=\boldsymbol{{cota}}−\boldsymbol{{cot}}\mathrm{2}\boldsymbol{{a}} \\ $$$$\boldsymbol{{cosec}}\mathrm{2}^{\mathrm{2}} \boldsymbol{{a}}=\boldsymbol{{cot}}\mathrm{2}\boldsymbol{{a}}−\boldsymbol{{cot}}\mathrm{2}^{\mathrm{2}} \boldsymbol{{a}} \\ $$$$.... \\ $$$$.... \\ $$$$\boldsymbol{{cosec}}\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{a}}=\boldsymbol{{cot}}\mathrm{2}^{\boldsymbol{{n}}−\mathrm{2}} \boldsymbol{{a}}−\boldsymbol{{cot}}\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{a}} \\ $$$$\boldsymbol{{add}}\:\boldsymbol{{them}} \\ $$$$\boldsymbol{{S}}={cot}\frac{{a}}{\mathrm{2}}−{cot}\mathrm{2}^{{n}−\mathrm{1}} {a} \\ $$$${S}={cot}\mathrm{1}^{{o}} −{cot}\left(\mathrm{2}^{\mathrm{1028}} \right)^{{o}} \\ $$