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

Show that 180°−2cos^(−1) (2/(√(13)))=sin^(−1) (((12)/(13)))

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

Answered by Yozzii last updated on 18/Jun/16

π−2acos(2/(√(13)))=asin(12/13) −−−−−−−−−−−−−−−−−−−− Let u=π−2acos(2/(√(13))). ∴sinu=sin(π−2acos(2/(√(13))) sinu=sinπcos(2acos(2/(√(13))))−sin(2acos(2/(√(13)))cosπ sinu=sin(2cos^(−1) (2/(√(13)))) sinu=2sin(cos^(−1) (2/(√(13))))cos(cos^(−1) (2/(√(13)))) sinu=(4/(√(13)))sin(cos^(−1) (2/(√(13)))) Let r=cos^(−1) (2/(√(13)))⇒cosr=(2/(√(13))) ⇒sinr=((√(13−4))/(√(13)))=(3/(√(13))). ⇒r=sin^(−1) (3/(√(13))) as principle result. ∴sinu=(4/(√(13)))sin(sin^(−1) (3/(√(13)))) sinu=(4/(√(13)))×(3/(√(13)))=((12)/(13))⇒u=sin^(−1) ((12)/(13)) as principle result. or π−2cos^(−1) (2/(√(13)))=sin^(−1) ((12)/(13)) (shown).

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

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

THANKS !

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

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