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Question Number 116887 by ZiYangLee last updated on 07/Oct/20
Let k=sin 1°×sin 3°×sin 5°×…×sin 89° Find log_2 k^2 .
$$\mathrm{Let}\:{k}=\mathrm{sin}\:\mathrm{1}°×\mathrm{sin}\:\mathrm{3}°×\mathrm{sin}\:\mathrm{5}°×\ldots×\mathrm{sin}\:\mathrm{89}° \\ $$$$\mathrm{Find}\:\mathrm{log}_{\mathrm{2}} {k}^{\mathrm{2}} . \\ $$
Commented by MJS_new last updated on 07/Oct/20
answer is −89
$$\mathrm{answer}\:\mathrm{is}\:−\mathrm{89} \\ $$
Answered by TANMAY PANACEA last updated on 07/Oct/20
89=1+(n−1)2→n=45 middle term 23rd term=sin45 k=sin1sin3sin5....sin45sin47sin49...sin89 k=(sin1sin89)(sin3sin87)(sin5sin85)...(sin43sin47)sin45 k×2^(22) =(2sin1cos1)(2sin3cos3)(2sin5cos5)...(2sin43cos43)sin45 =(sin2)(sin6)(sin10)...(sin86)sin45 wait
$$\mathrm{89}=\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}\rightarrow{n}=\mathrm{45} \\ $$$${middle}\:{term}\:\mathrm{23}{rd}\:{term}={sin}\mathrm{45} \\ $$$${k}={sin}\mathrm{1}{sin}\mathrm{3}{sin}\mathrm{5}….{sin}\mathrm{45}{sin}\mathrm{47}{sin}\mathrm{49}…{sin}\mathrm{89} \\ $$$${k}=\left({sin}\mathrm{1}{sin}\mathrm{89}\right)\left({sin}\mathrm{3}{sin}\mathrm{87}\right)\left({sin}\mathrm{5}{sin}\mathrm{85}\right)…\left({sin}\mathrm{43}{sin}\mathrm{47}\right){sin}\mathrm{45} \\ $$$${k}×\mathrm{2}^{\mathrm{22}} \\ $$$$=\left(\mathrm{2}{sin}\mathrm{1}{cos}\mathrm{1}\right)\left(\mathrm{2}{sin}\mathrm{3}{cos}\mathrm{3}\right)\left(\mathrm{2}{sin}\mathrm{5}{cos}\mathrm{5}\right)…\left(\mathrm{2}{sin}\mathrm{43}{cos}\mathrm{43}\right){sin}\mathrm{45} \\ $$$$=\left({sin}\mathrm{2}\right)\left({sin}\mathrm{6}\right)\left({sin}\mathrm{10}\right)…\left({sin}\mathrm{86}\right){sin}\mathrm{45} \\ $$$${wait} \\ $$
Answered by Bird last updated on 08/Oct/20
k =sin((π/(180))).sin(((3π)/(180)))...sin(((89π)/(180))) =Π_(k=0) ^(44) sin((((2k+1)π)/(180))) log_2 k^2 =((lnk^2 )/(ln2)) =(2/(ln2))lnk =(2/(ln2))Σ_(n=0) ^(44) sin((((2n+1)π)/(180))) but Σ_(n=0) ^(44) sin((((2n+1)π)/(180))) =Im(Σ_(n=0) ^(44) e^((i(2n+1)π)/(180)) ) and Σ_(n=0) ^(44) e^((i(2n+1)π)/(180)) =e^((iπ)/(180)) Σ_(n=0) ^(44 ) (e^((iπ)/(90)) )^n =e^((iπ)/(180)) ×((1−(e^((iπ)/(90)) )^(45) )/(1−e^((iπ)/(90)) )) =e^((iπ)/(180)) ×((1−i)/(1−e^((iπ)/(90)) )) =(1−i)e^((iπ)/(180)) ×(1/(1−cos((π/(90)))−isin((π/(90))))) =(1−i)e^((iπ)/(180)) ×((1−cos((π/(90)))+isin((π/(90))))/((1+cos((π/(90))))^2 +sin^2 ((π/(90))))) now its eazy to extract Im(of this qusntity...)
$${k}\:={sin}\left(\frac{\pi}{\mathrm{180}}\right).{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{180}}\right)…{sin}\left(\frac{\mathrm{89}\pi}{\mathrm{180}}\right) \\ $$$$=\prod_{{k}=\mathrm{0}} ^{\mathrm{44}} \:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{180}}\right) \\ $$$${log}_{\mathrm{2}} {k}^{\mathrm{2}} \:=\frac{{lnk}^{\mathrm{2}} }{{ln}\mathrm{2}}\:=\frac{\mathrm{2}}{{ln}\mathrm{2}}{lnk} \\ $$$$=\frac{\mathrm{2}}{{ln}\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\mathrm{44}} \:{sin}\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{180}}\right) \\ $$$${but}\:\sum_{{n}=\mathrm{0}} ^{\mathrm{44}} \:{sin}\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{180}}\right) \\ $$$$={Im}\left(\sum_{{n}=\mathrm{0}} ^{\mathrm{44}} \:\:{e}^{\frac{{i}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{180}}} \right) \\ $$$${and} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\mathrm{44}} \:{e}^{\frac{{i}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{180}}} \:={e}^{\frac{{i}\pi}{\mathrm{180}}} \:\sum_{{n}=\mathrm{0}} ^{\mathrm{44}\:} \left({e}^{\frac{{i}\pi}{\mathrm{90}}} \right)^{{n}} \\ $$$$={e}^{\frac{{i}\pi}{\mathrm{180}}} ×\frac{\mathrm{1}−\left({e}^{\frac{{i}\pi}{\mathrm{90}}} \right)^{\mathrm{45}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{90}}} } \\ $$$$={e}^{\frac{{i}\pi}{\mathrm{180}}} ×\frac{\mathrm{1}−{i}}{\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{90}}} } \\ $$$$=\left(\mathrm{1}−{i}\right){e}^{\frac{{i}\pi}{\mathrm{180}}} ×\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{90}}\right)−{isin}\left(\frac{\pi}{\mathrm{90}}\right)} \\ $$$$=\left(\mathrm{1}−{i}\right){e}^{\frac{{i}\pi}{\mathrm{180}}} ×\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{90}}\right)+{isin}\left(\frac{\pi}{\mathrm{90}}\right)}{\left(\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{90}}\right)\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{90}}\right)} \\ $$$${now}\:{its}\:{eazy}\:{to}\:{extract}\:{Im}\left({of}\right. \\ $$$$\left.{this}\:{qusntity}…\right) \\ $$

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