tng-9-4sin-9-3-

Question Number 99869 by bachamohamed last updated on 23/Jun/20

$$\:\boldsymbol{{tng}}\frac{\boldsymbol{\pi}}{\mathrm{9}}\:\:+\:\mathrm{4}\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{9}}\:=\sqrt{\mathrm{3}} \\ $$

Answered by smridha last updated on 23/Jun/20

$$\frac{\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)+\mathrm{2}\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\frac{\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)+\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{18}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}\:\: \\ $$$$=\frac{\boldsymbol{{sin}}\left(\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{9}}\right)+\boldsymbol{{sin}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\frac{\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}{\boldsymbol{{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}=\sqrt{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 23/Jun/20

$${Taking}\:\frac{\pi}{\mathrm{9}}\:{as}\:\mathrm{20}° \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left[\mathrm{4}{sin}\mathrm{20}°{cos}\mathrm{20}°+{sin}\mathrm{20}°\right] \\ $$$$=\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left[\mathrm{2}{sin}\mathrm{40}°+{sin}\mathrm{20}°\right] \\ $$$$=\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left[{sin}\mathrm{40}°+{sin}\mathrm{80}°\right] \\ $$$$=\frac{\mathrm{1}}{{cos}\mathrm{20}°}\left[\mathrm{2}{sin}\mathrm{60}°{cos}\mathrm{20}°\right]=\sqrt{\mathrm{3}}\left[{proved}\right] \\ $$

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