# prove-that-tan-3pi-7-4sin-pi-7-7-

Question Number 69081 by myintkhaing1121960@gmail.com last updated on 19/Sep/19
$${prove}\:{that} \\$$$${tan}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\:−\mathrm{4}{sin}\:\frac{\pi}{\mathrm{7}}\:=\:\sqrt{\mathrm{7}}. \\$$
Answered by Tanmay chaudhury last updated on 19/Sep/19
$$\mathrm{7}{a}=\pi \\$$$${tan}\mathrm{7}{a}={tan}\pi=−\mathrm{0} \\$$$$\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} −{s}_{\mathrm{7}} }{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} −{s}_{\mathrm{6}} }=\mathrm{0} \\$$$$\mathrm{7}{c}_{\mathrm{1}} {tana}−\mathrm{7}{c}_{\mathrm{3}} {tan}^{\mathrm{3}} {a}+\mathrm{7}{c}_{\mathrm{5}} {tan}^{\mathrm{5}} {a}−\mathrm{7}{c}_{\mathrm{7}} {tan}^{\mathrm{7}} {a}=\mathrm{0} \\$$$$\mathrm{7}{tana}−\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}}{\mathrm{3}×\mathrm{2}}{tan}^{\mathrm{3}} {a}+\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}}{tan}^{\mathrm{5}} {a}−{tan}^{\mathrm{7}} {a}=\mathrm{0} \\$$$$\mathrm{7}{tana}−\mathrm{35}{tan}^{\mathrm{3}} {a}+\mathrm{21}{tan}^{\mathrm{5}} {a}−{tan}^{\mathrm{7}} {a}=\mathrm{0} \\$$$${tana}={tan}\frac{\pi}{\mathrm{7}}\neq\mathrm{0} \\$$$$\mathrm{7}−\mathrm{35}{tan}^{\mathrm{2}} {a}+\mathrm{21}{tan}^{\mathrm{4}} {a}−{tan}^{\mathrm{6}} {a}=\mathrm{0} \\$$$$\boldsymbol{{tan}}^{\mathrm{6}} \boldsymbol{{a}}−\mathrm{21}\boldsymbol{{tan}}^{\mathrm{4}} \boldsymbol{{a}}+\mathrm{35}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}−\mathrm{7}=\mathrm{0} \\$$$${from}\:{above}\:{eqn}\:{we}\:{have}\:{to}\:{find}\:{value}\:{of}\:{tana} \\$$$${tan}\mathrm{3}{a}=\frac{\mathrm{3}{tana}−{tan}^{\mathrm{3}} {a}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {a}} \\$$$${sina} \\$$$$=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{sec}^{\mathrm{2}} {a}}}\:\: \\$$$$=\sqrt{\frac{{tan}^{\mathrm{2}} {a}}{\mathrm{1}+{tan}^{\mathrm{2}} {a}}}\: \\$$$$\boldsymbol{{tan}}\mathrm{3}\boldsymbol{{a}}−\mathrm{4}\boldsymbol{{sina}} \\$$$$=\frac{\mathrm{3}\boldsymbol{{tana}}−\boldsymbol{{tan}}^{\mathrm{3}} \boldsymbol{{a}}}{\mathrm{1}−\mathrm{3}\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}}−\frac{\mathrm{4}\boldsymbol{{tana}}}{\:\sqrt{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{a}}}\:} \\$$$${wait}… \\$$$$\\$$$$\\$$
Answered by mind is power last updated on 19/Sep/19
$${we}\:{have}\: \\$$$$\underset{{i}=\mathrm{1}} {\overset{{m}} {\prod}}{tan}\left(\frac{{i}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{m}+\mathrm{1}} \\$$$${we}\:{try}\:{too}\:{applie}\:{this}\:{gor}\:{m}=\mathrm{3} \\$$$$\Rightarrow\mathrm{tan}\:\left(\frac{\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\sqrt{\mathrm{7}} \\$$$${we}\:{finish}\:{if}\:{and}\:{only}\:{if}\: \\$$$$\Rightarrow\mathrm{tan}\:\left(\frac{\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\mathrm{tan}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{4sin}\:\left(\frac{\pi}{\mathrm{7}}\right) \\$$$${let}\:{a}=\frac{\pi}{\mathrm{7}} \\$$$${we}\:{want}\:{too}\:{show} \\$$$$\mathrm{tan}\:\left({a}\right)\mathrm{tan}\:\left(\mathrm{2}{a}\right)\mathrm{tan}\:\left(\mathrm{3}{a}\right)=\mathrm{tan}\:\left(\mathrm{3}{a}\right)−\mathrm{4sin}\:\left({a}\right) \\$$$$\Leftrightarrow\frac{\mathrm{sin}\:\left({a}\right)\mathrm{sin}\:\left(\mathrm{2}{a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{2}{a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}=\frac{\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{4sin}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{3}{a}\right)} \\$$$$\Leftrightarrow\frac{\mathrm{sin}\:\left({a}\right).\mathrm{2sin}\:\left({a}\right)\mathrm{cos}\:\left({a}\right)\mathrm{sin}\:\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{2}{a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)}=\frac{\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2sin}\:\left({a}\right){cos}\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{3}{a}\right)} \\$$$$\Leftrightarrow\frac{\mathrm{2sin}\:^{\mathrm{2}} \left({a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{a}\right)}=\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2cos}\:\left(\mathrm{3}{a}\right)\mathrm{sin}\:\left({a}\right) \\$$$$\mathrm{sin}\:\left({x}\right)\mathrm{cos}\:\left({y}\right)=\frac{{sin}\left({x}+{y}\right)+{sin}\left({x}−{y}\right)}{\mathrm{2}}\Rightarrow\mathrm{2sin}\:\left({a}\right)\mathrm{cos}\:\left(\mathrm{3}{a}\right)={sin}\left(\mathrm{4}{a}\right)−{sin}\left(\mathrm{2}{a}\right) \\$$$$\mathrm{2}{sin}^{\mathrm{2}} \left({a}\right)=\mathrm{1}−{cos}\left(\mathrm{2}{a}\right) \\$$$$\Rightarrow \\$$$$\frac{\mathrm{2sin}\:^{\mathrm{2}} \left({a}\right)\mathrm{sin}\:\left(\mathrm{3}{a}\right)}{\mathrm{cos}\:\left(\mathrm{2}{a}\right)}=\mathrm{sin}\:\left(\mathrm{3}{a}\right)−\mathrm{2}.\mathrm{2cos}\:\left(\mathrm{3}{a}\right)\mathrm{sin}\:\left({a}\right) \\$$$$\Leftrightarrow\frac{{sin}\left(\mathrm{3}{a}\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{a}\right)\right)}{{cos}\left(\mathrm{2}{a}\right)}={sin}\left(\mathrm{3}{a}\right)−\mathrm{2}{sin}\left(\mathrm{4}{a}\right)+\mathrm{2}{sin}\left(\mathrm{2}{a}\right) \\$$$$\Leftrightarrow{sin}\left(\mathrm{3}{a}\right)−{sin}\left(\mathrm{3}{a}\right){cos}\left(\mathrm{2}{a}\right)={sin}\left(\mathrm{3}{a}\right){cos}\left(\mathrm{2}{a}\right)−\mathrm{2}{sin}\left(\mathrm{4}{a}\right){cos}\left(\mathrm{2}{a}\right)+\mathrm{2}{sin}\left(\mathrm{2}{a}\right){cos}\left(\mathrm{2}{a}\right) \\$$$$\Leftrightarrow{sin}\left(\mathrm{3}{a}\right)−\frac{{sin}\left(\mathrm{5}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{5}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left(\mathrm{6}{a}\right)−{sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{4}{a}\right) \\$$$${sin}\left(\mathrm{4}{a}\right)={sin}\left(\mathrm{3}{a}\right)….{a}=\frac{\pi}{\mathrm{7}} \\$$$${sin}\left(\mathrm{6}{a}\right)={sin}\left({a}\right) \\$$$${sin}\left(\mathrm{5}{a}\right)={sin}\left(\mathrm{2}{a}\right) \\$$$$\Rightarrow{sin}\left(\mathrm{3}{a}\right)−\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left({a}\right)−{sin}\left(\mathrm{2}{a}\right)+{sin}\left(\mathrm{3}{a}\right) \\$$$$\Rightarrow\mathrm{0}=\frac{{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)+{sin}\left(\mathrm{2}{a}\right)+{sin}\left({a}\right)}{\mathrm{2}}−{sin}\left({a}\right)−{sin}\left(\mathrm{2}{a}\right)−{sin}\left(\mathrm{3}{a}\right)+{sin}\left(\mathrm{3}{a}\right) \\$$$$\Leftrightarrow\mathrm{0}=\mathrm{0} \\$$$$\Rightarrow{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{7}}\right)={tan}\left(\frac{\pi}{\mathrm{7}}\right){tan}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right){tan}\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\mathrm{tan}\:\left(\frac{{i}\pi}{\mathrm{7}}\right)=\sqrt{\mathrm{7}} \\$$$$\\$$$$\\$$
Commented by MJS last updated on 19/Sep/19
$$\mathrm{great}! \\$$
Commented by mind is power last updated on 19/Sep/19
$${thank}\:{you} \\$$$$\\$$