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Question Number 119937 by bobhans last updated on 28/Oct/20
(i)((1/2)−cos (π/7))((1/2)−cos ((3π)/7))((1/2)−cos ((9π)/7))? (ii) ((√3)+tan 1°)((√3)+tan 2°)((√3)+tan 3°)×...×((√3)+tan 29°)?
$$\:\left({i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{7}}\right)? \\ $$$$\left({ii}\right)\:\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{1}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{2}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{3}°\right)×…×\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{29}°\right)? \\ $$
Answered by TANMAY PANACEA last updated on 28/Oct/20
ii) (√3) +tan1^o =tan60^o +tan1^o =((sin60cos1+cos6osin1)/(cos60cos1))=((sin(61))/(cos60cos1)) now p=((sin61×sin62×sin63×...×sin89)/((cos60)^(29) cos1cos2cos3...cos29)) =2^(29) =
$$\left.{ii}\right)\:\sqrt{\mathrm{3}}\:+{tan}\mathrm{1}^{{o}} \\ $$$$={tan}\mathrm{60}^{{o}} +{tan}\mathrm{1}^{{o}} \\ $$$$=\frac{{sin}\mathrm{60}{cos}\mathrm{1}+{cos}\mathrm{6}{osin}\mathrm{1}}{{cos}\mathrm{60}{cos}\mathrm{1}}=\frac{{sin}\left(\mathrm{61}\right)}{{cos}\mathrm{60}{cos}\mathrm{1}} \\ $$$$\boldsymbol{{now}} \\ $$$${p}=\frac{{sin}\mathrm{61}×{sin}\mathrm{62}×{sin}\mathrm{63}×…×{sin}\mathrm{89}}{\left({cos}\mathrm{60}\right)^{\mathrm{29}} \boldsymbol{{cos}}\mathrm{1}\boldsymbol{{cos}}\mathrm{2}{cos}\mathrm{3}…{cos}\mathrm{29}} \\ $$$$=\mathrm{2}^{\mathrm{29}} \\ $$$$ \\ $$$$= \\ $$$$ \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 28/Oct/20
7θ=π cos3θ=cos(π−4θ) 4cos^3 θ−3cosθ=−(2cos^2 2θ−1) 4cos^3 θ−3cosθ=−2(2cos^2 θ−1)^2 +1 4c^3 −3c=−2(4c^4 −4c^2 +1)+1 4c^3 −3c=−8c^4 +8c^2 −2+1 8c^4 +4c^3 −8c^2 −3c+1=0 ⇛8c^4 +8c^3 −4c^3 −4c^2 −4c^2 −4c+c+1=0 ⇛8c^3 (c+1)−4c^2 (c+1) −4c (c+1)+1 (c+1)=0 ⇛(c+1)(8c^3 −4c^2 −4c+1)=0 given problem (a−c_1 )(a−c_2 )(a−c_3 ) a^3 −a^2 (c_1 +c_2 +c_3 )+a(c_1 c_2 +c_2 c_3 +c_1 c_3 )−c_1 c_2 c_3 a=(1/2) c_1 =cos(π/(7 )) c_2 =cos((3π)/7) c_3 =cos((9π)/7) cos(π/7),cos((3π)/7)and cos((9π)/7) satisfy eqn 8c^3 −4c^2 −4c+1=0 So answer is c_1 +c_2 +c_3 =((−(−4))/8) c_1 c_2 +c_2 c_3 +c_1 c_3 =((−4)/8) c_1 c_2 c_3 =((−1)/8) so answer is a^3 −a^2 (c_1 +c_2 +c_3 )+a(c_1 c_2 +c_2 c_3 +c_1 c_3 )−c_1 c_2 c_3 =(1/8)−(1/4)((1/2))+(1/2)(((−1)/2))−(((−1)/8)) =(1/8)−(1/8)−(1/4)+(1/8) =((1−2)/8)=((−1)/8) pls chk
$$\mathrm{7}\theta=\pi \\ $$$${cos}\mathrm{3}\theta={cos}\left(\pi−\mathrm{4}\theta\right) \\ $$$$\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta=−\left(\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{1}\right) \\ $$$$\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta=−\mathrm{2}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}=−\mathrm{2}\left(\mathrm{4}{c}^{\mathrm{4}} −\mathrm{4}{c}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{1} \\ $$$$\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}=−\mathrm{8}{c}^{\mathrm{4}} +\mathrm{8}{c}^{\mathrm{2}} −\mathrm{2}+\mathrm{1} \\ $$$$\mathrm{8}{c}^{\mathrm{4}} +\mathrm{4}{c}^{\mathrm{3}} −\mathrm{8}{c}^{\mathrm{2}} −\mathrm{3}{c}+\mathrm{1}=\mathrm{0} \\ $$$$\Rrightarrow\mathrm{8}{c}^{\mathrm{4}} +\mathrm{8}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}+{c}+\mathrm{1}=\mathrm{0} \\ $$$$\Rrightarrow\mathrm{8}{c}^{\mathrm{3}} \left({c}+\mathrm{1}\right)−\mathrm{4}{c}^{\mathrm{2}} \:\:\left({c}+\mathrm{1}\right)\:−\mathrm{4}{c}\:\left({c}+\mathrm{1}\right)+\mathrm{1}\:\left({c}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rrightarrow\left({c}+\mathrm{1}\right)\left(\mathrm{8}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$${given}\:{problem} \\ $$$$\left({a}−{c}_{\mathrm{1}} \right)\left({a}−{c}_{\mathrm{2}} \right)\left({a}−{c}_{\mathrm{3}} \right) \\ $$$${a}^{\mathrm{3}} −{a}^{\mathrm{2}} \left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} +{c}_{\mathrm{3}} \right)+{a}\left({c}_{\mathrm{1}} {c}_{\mathrm{2}} +{c}_{\mathrm{2}} {c}_{\mathrm{3}} +{c}_{\mathrm{1}} {c}_{\mathrm{3}} \right)−{c}_{\mathrm{1}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:{c}_{\mathrm{1}} ={cos}\frac{\pi}{\mathrm{7}\:}\:\:{c}_{\mathrm{2}} ={cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\:\:\:{c}_{\mathrm{3}} ={cos}\frac{\mathrm{9}\pi}{\mathrm{7}} \\ $$$${cos}\frac{\pi}{\mathrm{7}},{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}{and}\:{cos}\frac{\mathrm{9}\pi}{\mathrm{7}}\:{satisfy}\:{eqn} \\ $$$$\mathrm{8}{c}^{\mathrm{3}} −\mathrm{4}{c}^{\mathrm{2}} −\mathrm{4}{c}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{S}}{o}\:{answer}\:{is} \\ $$$${c}_{\mathrm{1}} +{c}_{\mathrm{2}} +{c}_{\mathrm{3}} =\frac{−\left(−\mathrm{4}\right)}{\mathrm{8}} \\ $$$${c}_{\mathrm{1}} {c}_{\mathrm{2}} +{c}_{\mathrm{2}} {c}_{\mathrm{3}} +{c}_{\mathrm{1}} {c}_{\mathrm{3}} =\frac{−\mathrm{4}}{\mathrm{8}} \\ $$$${c}_{\mathrm{1}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} =\frac{−\mathrm{1}}{\mathrm{8}} \\ $$$${so}\:{answer}\:{is} \\ $$$${a}^{\mathrm{3}} −{a}^{\mathrm{2}} \left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} +{c}_{\mathrm{3}} \right)+{a}\left({c}_{\mathrm{1}} {c}_{\mathrm{2}} +{c}_{\mathrm{2}} {c}_{\mathrm{3}} +{c}_{\mathrm{1}} {c}_{\mathrm{3}} \right)−{c}_{\mathrm{1}} {c}_{\mathrm{2}} {c}_{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)−\left(\frac{−\mathrm{1}}{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}}{\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{8}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{chk}} \\ $$
Commented by Dwaipayan Shikari last updated on 28/Oct/20
Yes it is correct sir!
$${Yes}\:{it}\:{is}\:{correct}\:{sir}! \\ $$
Commented by TANMAY PANACEA last updated on 28/Oct/20
thank you
$${thank}\:{you} \\ $$
Answered by bemath last updated on 28/Oct/20
(ii) ((√3)+tan 1°)((√3)+tan 29°)=3+(√3)(tan 29°+tan 1°)+tan 29°.tan 1° = 3+(√3) (tan 30°)(1−tan 29°tan 1°)+tan 29°.tan 1° = 3+1 = 4 similarly ((√3)+tan 2°)((√3)+tan 28°)=4 therefore we have =(4)(4)×...(4)×((√3) +tan 15°) = 2^(28) ×((√3)+((1−(1/( (√3))))/(1+(1/( (√3)))))) = 2^(28) ×((√3)+(((√3)−1)/( (√3)+1))) = 2^(28) ×((√3)+((4−2(√3))/2))=2^(28) ×((√3)+2−(√3)) = 2^(29)
$$\left({ii}\right)\:\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{1}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{29}°\right)=\mathrm{3}+\sqrt{\mathrm{3}}\left(\mathrm{tan}\:\mathrm{29}°+\mathrm{tan}\:\mathrm{1}°\right)+\mathrm{tan}\:\mathrm{29}°.\mathrm{tan}\:\mathrm{1}° \\ $$$$\:=\:\mathrm{3}+\sqrt{\mathrm{3}}\:\left(\mathrm{tan}\:\mathrm{30}°\right)\left(\mathrm{1}−\mathrm{tan}\:\mathrm{29}°\mathrm{tan}\:\mathrm{1}°\right)+\mathrm{tan}\:\mathrm{29}°.\mathrm{tan}\:\mathrm{1}° \\ $$$$\:=\:\mathrm{3}+\mathrm{1}\:=\:\mathrm{4} \\ $$$${similarly}\:\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{2}°\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{28}°\right)=\mathrm{4} \\ $$$${therefore}\:{we}\:{have} \\ $$$$=\left(\mathrm{4}\right)\left(\mathrm{4}\right)×…\left(\mathrm{4}\right)×\left(\sqrt{\mathrm{3}}\:+\mathrm{tan}\:\mathrm{15}°\right) \\ $$$$=\:\mathrm{2}^{\mathrm{28}} ×\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right)\:=\:\mathrm{2}^{\mathrm{28}} ×\left(\sqrt{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\right) \\ $$$$=\:\mathrm{2}^{\mathrm{28}} ×\left(\sqrt{\mathrm{3}}+\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{28}} ×\left(\sqrt{\mathrm{3}}+\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$=\:\mathrm{2}^{\mathrm{29}} \\ $$

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