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Question Number 112067 by bemath last updated on 06/Sep/20
(√(bemath −−−−■■)) find the value of Π_(m=1) ^6 tan (((mπ)/7)).
$$\:\:\sqrt{{bemath}\:−−−−\blacksquare\blacksquare} \\ $$$${find}\:{the}\:{value}\:{of}\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{7}}\right).\: \\ $$
Commented by MJS_new last updated on 06/Sep/20
should be =0
$$\mathrm{should}\:\mathrm{be}\:=\mathrm{0} \\ $$
Commented by bemath last updated on 06/Sep/20
why sir ? nothing choice answer 0
$${why}\:{sir}\:?\:{nothing}\:{choice}\:{answer}\:\mathrm{0} \\ $$
Commented by MJS_new last updated on 06/Sep/20
sorry the sum is 0 but the product is −7 as Sir John Santu has shown
$$\mathrm{sorry}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{0}\:\mathrm{but}\:\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:−\mathrm{7}\:\mathrm{as} \\ $$$$\mathrm{Sir}\:\mathrm{John}\:\mathrm{Santu}\:\mathrm{has}\:\mathrm{shown} \\ $$
Commented by bemath last updated on 06/Sep/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by john santu last updated on 06/Sep/20
Π_(m=1) ^6 tan (((mπ)/7)) = ? we have sin (nx) = 2^(m−1) Π_(m=0) ^(n−1) sin (x+((mπ)/n)) when x→0 , S_n = Π_(m=1) ^(n−1) sin (((mπ)/n))=(n/2^(n−1) ) where n = 7 , thus S_7 = Π_(m=1) ^(7−1) sin(((mπ)/7))=(7/(64)) but we know relation cos x=sin ((π/2)−x) cos (x+((mπ)/n))=−sin (x−(π/2)+((mπ)/n)) Π_(m=0) ^(n−1) cos (x+((mπ)/n))=(−1)^n Π_(m=0) ^(n−1) sin (x−(π/2)+((mπ)/n)) = (((−1)^n )/2^(n−1) ).sin (nx−((nπ)/2)) = (((−1)^n )/2^(n−1) ). ((sin (nx−((nπ)/2)))/(cos x)) x→0 ; Q_n = Π_(m=1) ^(n−1) cos (((mπ)/n))=(((−1)^(n+1) )/2^(n−1) ).sin (((nπ)/2)) put n= 7 ; Q_7 =Π_(m=1) ^(7−1) cos (((mπ)/7))=(((−1)^8 )/2^6 ).sin (((7π)/2))=−(1/(64)) therefore Π_(m=1) ^6 tan (((mπ)/7)) = ((((7/(64))))/((−(1/(64)))))=−7
$$\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{7}}\right)\:=\:? \\ $$$${we}\:{have}\:\mathrm{sin}\:\left({nx}\right)\:=\:\mathrm{2}^{{m}−\mathrm{1}} \:\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{sin}\:\left({x}+\frac{{m}\pi}{{n}}\right) \\ $$$${when}\:{x}\rightarrow\mathrm{0}\:,\:{S}_{{n}} =\:\underset{{m}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{sin}\:\left(\frac{{m}\pi}{{n}}\right)=\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$${where}\:{n}\:=\:\mathrm{7}\:,\:{thus}\:{S}_{\mathrm{7}} =\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{7}−\mathrm{1}} {\prod}}\mathrm{sin}\left(\frac{{m}\pi}{\mathrm{7}}\right)=\frac{\mathrm{7}}{\mathrm{64}}\: \\ $$$${but}\:{we}\:{know}\:{relation}\:\mathrm{cos}\:{x}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\mathrm{cos}\:\left({x}+\frac{{m}\pi}{{n}}\right)=−\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{2}}+\frac{{m}\pi}{{n}}\right) \\ $$$$\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{cos}\:\left({x}+\frac{{m}\pi}{{n}}\right)=\left(−\mathrm{1}\right)^{{n}} \:\underset{{m}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{2}}+\frac{{m}\pi}{{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}−\mathrm{1}} }.\mathrm{sin}\:\left({nx}−\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}−\mathrm{1}} }.\:\frac{\mathrm{sin}\:\left({nx}−\frac{{n}\pi}{\mathrm{2}}\right)}{\mathrm{cos}\:{x}} \\ $$$${x}\rightarrow\mathrm{0}\:;\:{Q}_{{n}} =\:\underset{{m}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\mathrm{cos}\:\left(\frac{{m}\pi}{{n}}\right)=\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}^{{n}−\mathrm{1}} }.\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$${put}\:{n}=\:\mathrm{7}\:;\:{Q}_{\mathrm{7}} =\underset{{m}=\mathrm{1}} {\overset{\mathrm{7}−\mathrm{1}} {\prod}}\mathrm{cos}\:\left(\frac{{m}\pi}{\mathrm{7}}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{8}} }{\mathrm{2}^{\mathrm{6}} }.\mathrm{sin}\:\left(\frac{\mathrm{7}\pi}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{64}} \\ $$$${therefore}\:\underset{{m}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\mathrm{tan}\:\left(\frac{{m}\pi}{\mathrm{7}}\right)\:=\:\frac{\left(\frac{\mathrm{7}}{\mathrm{64}}\right)}{\left(−\frac{\mathrm{1}}{\mathrm{64}}\right)}=−\mathrm{7} \\ $$$$ \\ $$
Commented by bemath last updated on 06/Sep/20
thank you
$${thank}\:{you} \\ $$

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