Question Number 17435 by 786786AM last updated on 06/Jul/17
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{4}\:\mathrm{sin}\:\frac{\pi}{\mathrm{24}}\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:\:\mathrm{cos}\frac{\pi}{\mathrm{6}}. \\ $$
Answered by alex041103 last updated on 07/Jul/17
![Let A=4sin(π/(24))cos(π/(12))cos(π/6). Then we use sin2θ=2sinθcosθ : A=2(2sin(π/(24))cos(π/(24)))cos(π/(12))cos(π/6) (1/(cos(π/(24)))) =(2sin(π/(12))cos(π/(12)))cos(π/6) (1/(cos(π/(24)))) =(1/(2cos(π/(24)))) (2sin(π/6)cos(π/6)) =((sin(π/3))/(2cos(π/(24)))) We know that cos2θ=2cos^2 θ−1 ⇒cos^2 (θ/2)=((1+cosθ)/2) When θ∈[0,(π/2)], (θ/2)∈[0,(π/2)]⇒cos(θ/2)≧0 ⇒cos(θ/2)=(√((1+cosθ)/2)) From this we can evaluate cos(π/(24)) cos(π/(12))=(√((1+cos(π/6))/2)) = (√((1+((√3)/2))/2)) =(√((2+(√3))/4)) = ((√(2+(√3)))/2) cos(π/(24))=(√((1+cos(π/(12)))/2)) = (√((1+((√(2+(√3)))/2))/2)) =(√((2+(√(2+(√3))))/4))=((√(2+(√(2+(√3)))))/2) And now we solve for A: A=(((√3)/2)/(2((√(2+(√(2+(√3)))))/2)))=((√3)/(2(√(2+(√(2+(√3))))))) A=(1/2)(√(3/(2+(√(2+(√3)))))) A=4sin(π/(24))cos(π/(12))cos(π/6)=(1/2)(√(3/(2+(√(2+(√3))))))≈0.436749...](https://www.tinkutara.com/question/Q17565.png)
$$\mathrm{Let}\:{A}=\mathrm{4}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}. \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{use}\:{sin}\mathrm{2}\theta=\mathrm{2}{sin}\theta{cos}\theta\:: \\ $$$${A}=\mathrm{2}\left(\mathrm{2}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{24}}\right){cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}\:\frac{\mathrm{1}}{{cos}\frac{\pi}{\mathrm{24}}} \\ $$$$=\left(\mathrm{2}{sin}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{12}}\right){cos}\frac{\pi}{\mathrm{6}}\:\frac{\mathrm{1}}{{cos}\frac{\pi}{\mathrm{24}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}\frac{\pi}{\mathrm{24}}}\:\left(\mathrm{2}{sin}\frac{\pi}{\mathrm{6}}{cos}\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{{sin}\frac{\pi}{\mathrm{3}}}{\mathrm{2}{cos}\frac{\pi}{\mathrm{24}}} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:{cos}\mathrm{2}\theta=\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}+{cos}\theta}{\mathrm{2}} \\ $$$$\mathrm{When}\:\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right],\:\frac{\theta}{\mathrm{2}}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\Rightarrow{cos}\frac{\theta}{\mathrm{2}}\geqq\mathrm{0} \\ $$$$\Rightarrow{cos}\frac{\theta}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}+{cos}\theta}{\mathrm{2}}} \\ $$$$\mathrm{From}\:\mathrm{this}\:\mathrm{we}\:\mathrm{can}\:\mathrm{evaluate}\:{cos}\frac{\pi}{\mathrm{24}} \\ $$$${cos}\frac{\pi}{\mathrm{12}}=\sqrt{\frac{\mathrm{1}+{cos}\frac{\pi}{\mathrm{6}}}{\mathrm{2}}}\:=\:\sqrt{\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}} \\ $$$$=\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}}\:=\:\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$${cos}\frac{\pi}{\mathrm{24}}=\sqrt{\frac{\mathrm{1}+{cos}\frac{\pi}{\mathrm{12}}}{\mathrm{2}}}\:=\:\sqrt{\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}}}{\mathrm{2}}} \\ $$$$=\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}{\mathrm{2}} \\ $$$$\mathrm{And}\:\mathrm{now}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{for}\:{A}: \\ $$$${A}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{3}}{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}} \\ $$$${A}=\mathrm{4}{sin}\frac{\pi}{\mathrm{24}}{cos}\frac{\pi}{\mathrm{12}}{cos}\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{3}}{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}\approx\mathrm{0}.\mathrm{436749}… \\ $$
Commented by mrW1 last updated on 07/Jul/17
$$\mathrm{very}\:\mathrm{good}\:\mathrm{sir}! \\ $$
Commented by alex041103 last updated on 07/Jul/17
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{Q}:\mathrm{17514}.\:\mathrm{I}'\mathrm{ll} \\ $$$$\mathrm{post}\:\mathrm{the}\:\mathrm{hole}\:\mathrm{prove}\:\mathrm{at}\:\mathrm{some}\:\mathrm{point}. \\ $$