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Question Number 112067 by bemath last updated on 06/Sep/20

$$\sqrt{bemath----◼◼}$$$$findthevalueof\underset{m=1}{\prod ^6}\tan \left(\frac{m\pi }{7}\right).$$

Commented by MJS_new last updated on 06/Sep/20

$$\mathrm{should}\mathrm{be}=0$$

Commented by bemath last updated on 06/Sep/20

$$whysir?nothingchoiceanswer0$$

Commented by MJS_new last updated on 06/Sep/20

$$\mathrm{sorry}\mathrm{the}\mathrm{sum}\mathrm{is}0\mathrm{but}\mathrm{the}\mathrm{product}\mathrm{is}-7\mathrm{as}$$$$\mathrm{Sir}\mathrm{John}\mathrm{Santu}\mathrm{has}\mathrm{shown}$$

Commented by bemath last updated on 06/Sep/20

$$thankyousir$$

Answered by john santu last updated on 06/Sep/20

$$\underset{m=1}{\prod ^6}\tan \left(\frac{m\pi }{7}\right)=?$$$$wehave\sin \left(nx\right)=2^{m-1}\underset{m=0}{\prod ^{n-1}}\sin (x+\frac{m\pi }{n})$$$$whenx\to 0,S_n=\underset{m=1}{\prod ^{n-1}}\sin \left(\frac{m\pi }{n}\right)=\frac{n}{2^{n-1}}$$$$wheren=7,thus{S}_7=\underset{m=1}{\prod ^{7-1}}\sin \left(\frac{m\pi }{7}\right)=\frac{7}{64}$$$$butweknowrelation\cos x=\sin (\frac{\pi }{2}-x)$$$$\cos (x+\frac{m\pi }{n})=-\sin (x-\frac{\pi }{2}+\frac{m\pi }{n})$$$$\underset{m=0}{\prod ^{n-1}}\cos (x+\frac{m\pi }{n})={(-1)}^n\underset{m=0}{\prod ^{n-1}}\sin (x-\frac{\pi }{2}+\frac{m\pi }{n})$$$$=\frac{{(-1)}^n}{2^{n-1}}.\sin (nx-\frac{n\pi }{2})$$$$=\frac{{(-1)}^n}{2^{n-1}}.\frac{\sin (nx-\frac{n\pi }{2})}{\cos x}$$$$x\to 0;Q_n=\underset{m=1}{\prod ^{n-1}}\cos \left(\frac{m\pi }{n}\right)=\frac{{(-1)}^{n+1}}{2^{n-1}}.\sin \left(\frac{n\pi }{2}\right)$$$$putn=7;Q_7=\underset{m=1}{\prod ^{7-1}}\cos \left(\frac{m\pi }{7}\right)=\frac{{(-1)}^8}{2^6}.\sin \left(\frac{7\pi }{2}\right)=-\frac{1}{64}$$$$therefore\underset{m=1}{\prod ^6}\tan \left(\frac{m\pi }{7}\right)=\frac{\left(\frac{7}{64}\right)}{(-\frac{1}{64})}=-7$$$$$$

Commented by bemath last updated on 06/Sep/20

$$thankyou$$

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