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Question Number 95326 by bobhans last updated on 24/May/20

sin 72^o = p(√3) cos 48^o find tan 12^o ?

$$\mathrm{sin}\:\mathrm{72}^{\mathrm{o}} \:=\:\mathrm{p}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{48}^{\mathrm{o}} \\ $$$$\mathrm{find}\:\mathrm{tan}\:\mathrm{12}^{\mathrm{o}} \:? \\ $$

Commented by mr W last updated on 24/May/20

sin 72°=sin (60°+12°)=(1/2)((√3)cos 12°+sin 12°) cos 48°=cos (60°−12°)=(1/2)(cos 12°+(√3)sin 12°) sin 72^o = p(√3) cos 48^o (1/2)((√3)cos 12°+sin 12°)=p(√3)(1/2)(cos 12°+(√3)sin 12°) (√3)+tan 12°=p(√3)(1+(√3)tan 12°) (√3)(1−p)=(3p−1)tan 12° ⇒tan 12°=(((1−p)(√3))/(3p−1))

$$\mathrm{sin}\:\mathrm{72}°=\mathrm{sin}\:\left(\mathrm{60}°+\mathrm{12}°\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{12}°+\mathrm{sin}\:\mathrm{12}°\right) \\ $$$$\mathrm{cos}\:\mathrm{48}°=\mathrm{cos}\:\left(\mathrm{60}°−\mathrm{12}°\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{12}°+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{12}°\right) \\ $$$$\mathrm{sin}\:\mathrm{72}^{\mathrm{o}} \:=\:\mathrm{p}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{48}^{\mathrm{o}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{12}°+\mathrm{sin}\:\mathrm{12}°\right)={p}\sqrt{\mathrm{3}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\mathrm{12}°+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{12}°\right) \\ $$$$\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{12}°={p}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{12}°\right) \\ $$$$\sqrt{\mathrm{3}}\left(\mathrm{1}−{p}\right)=\left(\mathrm{3}{p}−\mathrm{1}\right)\mathrm{tan}\:\mathrm{12}° \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{12}°=\frac{\left(\mathrm{1}−{p}\right)\sqrt{\mathrm{3}}}{\mathrm{3}{p}−\mathrm{1}} \\ $$

Commented by bobhans last updated on 24/May/20

yes...

$$\mathrm{yes}... \\ $$

Commented by bobhans last updated on 24/May/20

how with qn 95260

$$\mathrm{how}\:\mathrm{with}\:\mathrm{qn}\:\mathrm{95260} \\ $$

Commented by bobhans last updated on 24/May/20

sir jagoll asking to you

$$\mathrm{sir}\:\mathrm{jagoll}\:\mathrm{asking}\:\mathrm{to}\:\mathrm{you} \\ $$

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