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Question Number 51356 by rahul 19 last updated on 26/Dec/18

Evaluate : tan {(1/2)cos^(−1) ((√5)/3)} ?

$${Evaluate}\:: \\ $$$$\mathrm{tan}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\right\}\:? \\ $$

Commented by rahul 19 last updated on 26/Dec/18

Ans→ ((3−(√5))/2).

$${Ans}\rightarrow\:\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}. \\ $$

Answered by mr W last updated on 26/Dec/18

let c=cos^(−1) ((√5)/3) ⇒cos c=((√5)/3) tan {(1/2)cos^(−1) ((√5)/3)}=tan (c/2)=(√((1−cos c)/(1+cos c))) =(√((1−((√5)/3))/(1+((√5)/3))))=(√((3−(√5))/(3+(√5))))=(√(((3−(√5))^2 )/((3+(√5))(3−(√5))))) =((3−(√5))/2)

$${let}\:{c}=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{cos}\:{c}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\right\}=\mathrm{tan}\:\frac{{c}}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:{c}}{\mathrm{1}+\mathrm{cos}\:{c}}} \\ $$$$=\sqrt{\frac{\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}}{\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}}}=\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}}=\sqrt{\frac{\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)}} \\ $$$$=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by rahul 19 last updated on 26/Dec/18

thank you sir! Definitely shorter than the previous one!:)

$${thank}\:{you}\:{sir}! \\ $$$${Definitely}\:{shorter}\:{than}\:{the}\:{previous} \\ $$$$\left.{one}!:\right) \\ $$

Answered by ajfour last updated on 26/Dec/18

tan [(1/2)(2θ)]=tan [(1/2)sin^(−1) (2/3)] ⇒ sin 2θ = (2/3) = ((2t)/(1+t^2 )) ∀ t =tan θ ⇒ t^2 −3t+1 = 0 t = ((3±(√(9−4)))/2) = ((3±(√5))/2) .

$$\mathrm{tan}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\theta\right)\right]=\mathrm{tan}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}\right] \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:\:\forall\:{t}\:=\mathrm{tan}\:\theta \\ $$$$\:\Rightarrow\:\:{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{t}\:=\:\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:. \\ $$

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