Show-that-180-2cos-1-2-13-sin-1-12-13-

Question Number 6183 by Rasheed Soomro last updated on 17/Jun/16

$${Show}\:{that} \\ $$$$\mathrm{180}°−\mathrm{2cos}^{−\mathrm{1}} \:\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$

Answered by Yozzii last updated on 18/Jun/16

$$\pi−\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right)={asin}\left(\mathrm{12}/\mathrm{13}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{u}=\pi−\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right). \\ $$$$\therefore{sinu}={sin}\left(\pi−\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right)\right. \\ $$$${sinu}={sin}\pi{cos}\left(\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right)\right)−{sin}\left(\mathrm{2}{acos}\left(\mathrm{2}/\sqrt{\mathrm{13}}\right){cos}\pi\right. \\ $$$${sinu}={sin}\left(\mathrm{2}{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\right) \\ $$$${sinu}=\mathrm{2}{sin}\left({cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\right){cos}\left({cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\right) \\ $$$${sinu}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{13}}}{sin}\left({cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\right) \\ $$$${Let}\:{r}={cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\Rightarrow{cosr}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}\:\:\: \\ $$$$\Rightarrow{sinr}=\frac{\sqrt{\mathrm{13}−\mathrm{4}}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}. \\ $$$$\Rightarrow{r}={sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\:{as}\:{principle}\:{result}. \\ $$$$\therefore{sinu}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{13}}}{sin}\left({sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\right) \\ $$$${sinu}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{13}}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{12}}{\mathrm{13}}\Rightarrow{u}={sin}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}\:\:{as}\:{principle}\:{result}. \\ $$$${or}\:\pi−\mathrm{2}{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}={sin}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}\:\left({shown}\right). \\ $$

Commented by Rasheed Soomro last updated on 18/Jun/16

$$\mathscr{THANKS}\:! \\ $$

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