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Question Number 69081 by myintkhaing1121960@gmail.com last updated on 19/Sep/19

prove that tan ((3π)/7) −4sin (π/7) = (√7).

$${prove}\:{that} \\ $$$${tan}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:−\mathrm{4}{sin}\:\frac{\pi}{\mathrm{7}}\:=\:\sqrt{\mathrm{7}}. \\ $$

Answered by Tanmay chaudhury last updated on 19/Sep/19

7a=π tan7a=tanπ=−0 ((s_1 −s_3 +s_5 −s_7 )/(1−s_2 +s_4 −s_6 ))=0 7c_1 tana−7c_3 tan^3 a+7c_5 tan^5 a−7c_7 tan^7 a=0 7tana−((7×6×5)/(3×2))tan^3 a+((7×6)/2)tan^5 a−tan^7 a=0 7tana−35tan^3 a+21tan^5 a−tan^7 a=0 tana=tan(π/7)≠0 7−35tan^2 a+21tan^4 a−tan^6 a=0 tan^6 a−21tan^4 a+35tan^2 a−7=0 from above eqn we have to find value of tana tan3a=((3tana−tan^3 a)/(1−3tan^2 a)) sina =(√(1−(1/(sec^2 a)))) =(√((tan^2 a)/(1+tan^2 a))) tan3a−4sina =((3tana−tan^3 a)/(1−3tan^2 a))−((4tana)/((√(1+tan^2 a)) )) wait...

$$\mathrm{7}{a}=\pi \\ $$$${tan}\mathrm{7}{a}={tan}\pi=−\mathrm{0} \\ $$$$\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} −{s}_{\mathrm{7}} }{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} −{s}_{\mathrm{6}} }=\mathrm{0} \\ $$$$\mathrm{7}{c}_{\mathrm{1}} {tana}−\mathrm{7}{c}_{\mathrm{3}} {tan}^{\mathrm{3}} {a}+\mathrm{7}{c}_{\mathrm{5}} {tan}^{\mathrm{5}} {a}−\mathrm{7}{c}_{\mathrm{7}} {tan}^{\mathrm{7}} {a}=\mathrm{0} \\ $$$$\mathrm{7}{tana}−\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{3}×\mathrm{2}}{tan}^{\mathrm{3}} {a}+\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}}{tan}^{\mathrm{5}} {a}−{tan}^{\mathrm{7}} {a}=\mathrm{0} \\ $$$$\mathrm{7}{tana}−\mathrm{35}{tan}^{\mathrm{3}} {a}+\mathrm{21}{tan}^{\mathrm{5}} {a}−{tan}^{\mathrm{7}} {a}=\mathrm{0} \\ $$$${tana}={tan}\frac{\pi}{\mathrm{7}}\neq\mathrm{0} \\ $$$$\mathrm{7}−\mathrm{35}{tan}^{\mathrm{2}} {a}+\mathrm{21}{tan}^{\mathrm{4}} {a}−{tan}^{\mathrm{6}} {a}=\mathrm{0} \\ $$$$\boldsymbol{{tan}}^{\mathrm{6}} \boldsymbol{{a}}−\mathrm{21}\boldsymbol{{tan}}^{\mathrm{4}} \boldsymbol{{a}}+\mathrm{35}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}−\mathrm{7}=\mathrm{0} \\ $$$${from}\:{above}\:{eqn}\:{we}\:{have}\:{to}\:{find}\:{value}\:{of}\:{tana} \\ $$$${tan}\mathrm{3}{a}=\frac{\mathrm{3}{tana}−{tan}^{\mathrm{3}} {a}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {a}} \\ $$$${sina} \\ $$$$=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{sec}^{\mathrm{2}} {a}}}\:\: \\ $$$$=\sqrt{\frac{{tan}^{\mathrm{2}} {a}}{\mathrm{1}+{tan}^{\mathrm{2}} {a}}}\: \\ $$$$\boldsymbol{{tan}}\mathrm{3}\boldsymbol{{a}}−\mathrm{4}\boldsymbol{{sina}} \\ $$$$=\frac{\mathrm{3}\boldsymbol{{tana}}−\boldsymbol{{tan}}^{\mathrm{3}} \boldsymbol{{a}}}{\mathrm{1}−\mathrm{3}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}}−\frac{\mathrm{4}\boldsymbol{{tana}}}{\sqrt{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}}\:} \\ $$$${wait}... \\ $$$$ \\ $$$$ \\ $$

Answered by mind is power last updated on 19/Sep/19

we have Π_(i=1) ^m tan(((iπ)/(2m+1)))=(√(2m+1)) we try too applie this gor m=3 ⇒tan ((π/7))tan (((2π)/7))tan (((3π)/7))=(√7) we finish if and only if ⇒tan ((π/7))tan (((2π)/7))tan (((3π)/7))=tan (((3π)/7))−4sin ((π/7)) let a=(π/7) we want too show tan (a)tan (2a)tan (3a)=tan (3a)−4sin (a) ⇔((sin (a)sin (2a)sin (3a))/(cos (a)cos (2a)cos (3a)))=((sin (3a)−4sin (a)cos (3a))/(cos (3a))) ⇔((sin (a).2sin (a)cos (a)sin (3a))/(cos (a)cos (2a)cos (3a)))=((sin (3a)−2.2sin (a)cos(3a))/(cos (3a))) ⇔((2sin^2 (a)sin (3a))/(cos (2a)))=sin (3a)−2.2cos (3a)sin (a) sin (x)cos (y)=((sin(x+y)+sin(x−y))/2)⇒2sin (a)cos (3a)=sin(4a)−sin(2a) 2sin^2 (a)=1−cos(2a) ⇒ ((2sin^2 (a)sin (3a))/(cos (2a)))=sin (3a)−2.2cos (3a)sin (a) ⇔((sin(3a)(1−cos(2a)))/(cos(2a)))=sin(3a)−2sin(4a)+2sin(2a) ⇔sin(3a)−sin(3a)cos(2a)=sin(3a)cos(2a)−2sin(4a)cos(2a)+2sin(2a)cos(2a) ⇔sin(3a)−((sin(5a)+sin(a))/2)=((sin(5a)+sin(a))/2)−sin(6a)−sin(2a)+sin(4a) sin(4a)=sin(3a)....a=(π/7) sin(6a)=sin(a) sin(5a)=sin(2a) ⇒sin(3a)−((sin(2a)+sin(a))/2)=((sin(2a)+sin(a))/2)−sin(a)−sin(2a)+sin(3a) ⇒0=((sin(2a)+sin(a)+sin(2a)+sin(a))/2)−sin(a)−sin(2a)−sin(3a)+sin(3a) ⇔0=0 ⇒tan(((3π)/7))−4sin((π/7))=tan((π/7))tan(((2π)/7))tan(((3π)/7))=Π_(i=1) ^3 tan (((iπ)/7))=(√7)

$${we}\:{have}\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{m}} {\prod}}{tan}\left(\frac{{i}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{m}+\mathrm{1}} \\ $$$${we}\:{try}\:{too}\:{applie}\:{this}\:{gor}\:{m}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{tan}\:\left(\frac{\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\sqrt{\mathrm{7}} \\ $$$${we}\:{finish}\:{if}\:{and}\:{only}\:{if}\: \\ $$$$\Rightarrow\mathrm{tan}\:\left(\frac{\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{4sin}\:\left(\frac{\pi}{\mathrm{7}}\right) \\ $$$${let}\:{a}=\frac{\pi}{\mathrm{7}} \\ $$$${we}\:{want}\:{too}\:{show} \\ $$$$\mathrm{tan}\:\left({a}\right)\mathrm{tan}\:\left(\mathrm{2}{a}\right)\mathrm{tan}\:\left(\mathrm{3}{a}\right)=\mathrm{tan}\:\left(\mathrm{3}{a}\right)−\mathrm{4sin}\:\left({a}\right) \\ $$$$\Leftrightarrow\frac{\mathrm{sin}\:\left({a}\right)\mathrm{sin}\:\left(\mathrm{2}{a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{2}{a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}=\frac{\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{4sin}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{3}{a}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{sin}\:\left({a}\right).\mathrm{2sin}\:\left({a}\right)\mathrm{cos}\:\left({a}\right)\mathrm{sin}\:\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{2}{a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}=\frac{\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2sin}\:\left({a}\right){cos}\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{3}{a}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{2sin}\:^{\mathrm{2}} \left({a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{a}\right)}=\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2cos}\:\left(\mathrm{3}{a}\right)\mathrm{sin}\:\left({a}\right) \\ $$$$\mathrm{sin}\:\left({x}\right)\mathrm{cos}\:\left({y}\right)=\frac{{sin}\left({x}+{y}\right)+{sin}\left({x}−{y}\right)}{\mathrm{2}}\Rightarrow\mathrm{2sin}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)={sin}\left(\mathrm{4}{a}\right)−{sin}\left(\mathrm{2}{a}\right) \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} \left({a}\right)=\mathrm{1}−{cos}\left(\mathrm{2}{a}\right) \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{2sin}\:^{\mathrm{2}} \left({a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{a}\right)}=\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2cos}\:\left(\mathrm{3}{a}\right)\mathrm{sin}\:\left({a}\right) \\ $$$$\Leftrightarrow\frac{{sin}\left(\mathrm{3}{a}\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{a}\right)\right)}{{cos}\left(\mathrm{2}{a}\right)}={sin}\left(\mathrm{3}{a}\right)−\mathrm{2}{sin}\left(\mathrm{4}{a}\right)+\mathrm{2}{sin}\left(\mathrm{2}{a}\right) \\ $$$$\Leftrightarrow{sin}\left(\mathrm{3}{a}\right)−{sin}\left(\mathrm{3}{a}\right){cos}\left(\mathrm{2}{a}\right)={sin}\left(\mathrm{3}{a}\right){cos}\left(\mathrm{2}{a}\right)−\mathrm{2}{sin}\left(\mathrm{4}{a}\right){cos}\left(\mathrm{2}{a}\right)+\mathrm{2}{sin}\left(\mathrm{2}{a}\right){cos}\left(\mathrm{2}{a}\right) \\ $$$$\Leftrightarrow{sin}\left(\mathrm{3}{a}\right)−\frac{{sin}\left(\mathrm{5}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{5}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left(\mathrm{6}{a}\right)−{sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{4}{a}\right) \\ $$$${sin}\left(\mathrm{4}{a}\right)={sin}\left(\mathrm{3}{a}\right)....{a}=\frac{\pi}{\mathrm{7}} \\ $$$${sin}\left(\mathrm{6}{a}\right)={sin}\left({a}\right) \\ $$$${sin}\left(\mathrm{5}{a}\right)={sin}\left(\mathrm{2}{a}\right) \\ $$$$\Rightarrow{sin}\left(\mathrm{3}{a}\right)−\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left({a}\right)−{sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{3}{a}\right) \\ $$$$\Rightarrow\mathrm{0}=\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)+{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left({a}\right)−{sin}\left(\mathrm{2}{a}\right)−{sin}\left(\mathrm{3}{a}\right)+{sin}\left(\mathrm{3}{a}\right) \\ $$$$\Leftrightarrow\mathrm{0}=\mathrm{0} \\ $$$$\Rightarrow{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{7}}\right)={tan}\left(\frac{\pi}{\mathrm{7}}\right){tan}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){tan}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\mathrm{tan}\:\left(\frac{{i}\pi}{\mathrm{7}}\right)=\sqrt{\mathrm{7}} \\ $$$$ \\ $$$$ \\ $$

Commented by MJS last updated on 19/Sep/19

great!

$$\mathrm{great}! \\ $$

Commented by mind is power last updated on 19/Sep/19

thank you

$${thank}\:{you} \\ $$$$ \\ $$

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