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Question-189125

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Question Number 189125 by Rupesh123 last updated on 12/Mar/23
Answered by cortano12 last updated on 12/Mar/23
cos 18°=2cos^2 9°−1 cos^2 9°=((1+cos 18°)/2) cos 18°=(√(1−sin^2 18°)) =(√(1−((((√5)−1)/4))^2 )) =((√(16−(6−2(√5))))/4)=((√(10+2(√5)))/4)
$$\:\mathrm{cos}\:\mathrm{18}°=\mathrm{2cos}^{\mathrm{2}} \:\mathrm{9}°−\mathrm{1} \\ $$$$\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{9}°=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{18}°}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{18}°=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{18}°} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{16}−\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 13/Mar/23
Excellent!
Answered by BaliramKumar last updated on 12/Mar/23
cos18° = ((√(10+2(√(5 ))))/4) 2cos^2 9° − 1 = ((√(10+2(√5)))/4) 8cos^2 9° − 4 = (√(10+2(√5))) 8cos^2 9° = 4 + (√(10+2(√5))) 2(√2)cos9° = (√(4+(√(10+2(√5))))) 2(√2)cos9° = (√(((4+(√(16−10−2(√5))))/2) )) + (√((4−(√(16−10−2(√5))))/2)) 2(√2)cos9° = (√(((4+(√(6−2(√5))))/2) )) + (√((4−(√(6−2(√5))))/2)) 2(√2)cos9° = (√(((4+ (√5)−1)/2) )) + (√((4−((√5)−1))/2)) 2(√2)cos9° = (√(((3+ (√5))/2) )) + (√((5−(√5))/2)) cos9° = ((√(3+(√5)))/4) + ((√(5−(√5)))/4) = ((2(√(3+(√5) ))+ 2(√(5−(√5))))/8) cos9° = (((√2)(√(6+2(√5) ))+ 2(√(5−(√5))))/8) cos9° = (((√2)(1+(√5))+ 2(√(5−(√5))))/8) cos9° = (((√2) + (√(10))+ 2(√(5−(√5))))/8) (√(a±(√b))) = (√((a+(√(a^2 −b)))/2)) ± (√((a−(√(a^2 −b)))/2))
$$ \\ $$$${cos}\mathrm{18}°\:=\:\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}\:}}}{\mathrm{4}} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \mathrm{9}°\:−\:\mathrm{1}\:=\:\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\: \\ $$$$\mathrm{8}{cos}^{\mathrm{2}} \mathrm{9}°\:−\:\mathrm{4}\:=\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\mathrm{8}{cos}^{\mathrm{2}} \mathrm{9}°\:=\:\mathrm{4}\:+\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{cos}\mathrm{9}°\:=\:\sqrt{\mathrm{4}+\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{cos}\mathrm{9}°\:=\:\sqrt{\frac{\mathrm{4}+\sqrt{\mathrm{16}−\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\:}\:+\:\sqrt{\frac{\mathrm{4}−\sqrt{\mathrm{16}−\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{cos}\mathrm{9}°\:=\:\sqrt{\frac{\mathrm{4}+\sqrt{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\:}\:+\:\sqrt{\frac{\mathrm{4}−\sqrt{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{cos}\mathrm{9}°\:=\:\sqrt{\frac{\mathrm{4}+\:\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:}\:+\:\sqrt{\frac{\mathrm{4}−\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{2}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}{cos}\mathrm{9}°\:=\:\sqrt{\frac{\mathrm{3}+\:\sqrt{\mathrm{5}}}{\mathrm{2}}\:}\:+\:\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$${cos}\mathrm{9}°\:=\:\frac{\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}}{\mathrm{4}}\:+\:\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{4}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}\:}+\:\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{8}} \\ $$$${cos}\mathrm{9}°\:=\:\frac{\sqrt{\mathrm{2}}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:}+\:\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{8}} \\ $$$${cos}\mathrm{9}°\:=\:\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)+\:\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{8}} \\ $$$${cos}\mathrm{9}°\:=\:\frac{\sqrt{\mathrm{2}}\:+\:\sqrt{\mathrm{10}}+\:\mathrm{2}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{8}} \\ $$$$ \\ $$$$ \\ $$$$\sqrt{{a}\pm\sqrt{{b}}}\:=\:\sqrt{\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}}\:\pm\:\sqrt{\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 13/Mar/23
Excellent!

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