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Question Number 102587 by 1549442205 last updated on 10/Jul/20

Prove that cot7.5°=(√2)+(√3)+(√4)+(√6)

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cot7}.\mathrm{5}°=\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{6}} \\ $$

Commented by PRITHWISH SEN 2 last updated on 10/Jul/20

a different approach cot θ=((cos θ)/(sin θ)) = ((2cos 7θcos θ)/(2cos 7θsin θ)) = ((cos 8θ+cos 6θ)/(sin 8θ−sin 6θ)) put θ=7∙5 cot 7∙5= ((cos 60+cos 45)/(sin 60−sin 45))=(((1+(√2)))/(((√3)−(√2))))=(1+(√2))((√3)+(√2)) =(√2)+ (√3) +(√4) +(√6)

$$\mathrm{a}\:\mathrm{different}\:\mathrm{approach} \\ $$$$\mathrm{cot}\:\theta=\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{2cos}\:\mathrm{7}\theta\mathrm{cos}\:\theta}{\mathrm{2cos}\:\mathrm{7}\theta\mathrm{sin}\:\theta} \\ $$$$=\:\frac{\mathrm{cos}\:\mathrm{8}\theta+\mathrm{cos}\:\mathrm{6}\theta}{\mathrm{sin}\:\mathrm{8}\theta−\mathrm{sin}\:\mathrm{6}\theta} \\ $$$$\mathrm{put}\:\theta=\mathrm{7}\centerdot\mathrm{5} \\ $$$$\mathrm{cot}\:\mathrm{7}\centerdot\mathrm{5}=\:\frac{\mathrm{cos}\:\mathrm{60}+\mathrm{cos}\:\mathrm{45}}{\mathrm{sin}\:\mathrm{60}−\mathrm{sin}\:\mathrm{45}}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right) \\ $$$$=\sqrt{\mathrm{2}}+\:\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:+\sqrt{\mathrm{6}} \\ $$

Answered by Dwaipayan Shikari last updated on 10/Jul/20

Method 1 ) cot7.5°=((cos7.5°)/(sin7.5))=((2cos^2 7.5°)/(sin15°))=((1+cos15°)/(sin15°))=((2cos15°+1+cos30°)/(sin30°)) cos15=(((√3)+1)/(2(√2))) =2((((√3)+1)/(√2))+1+(√3).(1/2)) =2((((√6)+(√2)+(√4)+(√3))/2)) so cot7.5°=(√2)+(√3)+(√4)+(√6) ( proved) Method 2 cos(π/(24))=(√((1/2)(1+cos(π/(12)))))=(√((1/2)(1+(((√3)+1)/(2(√2))))))=(1/2)(√((2(√2)+(√3)+1)/(√2))) (1/(2(√2)))(√(4+(√6)+(√2))) sin(π/(24))=(1/(2(√2))).(√(4−(√6)−(√2))) cot(π/(24))=(√((4+(√6)+(√2))/(4−(√6)−(√2))))=((4+(√6)+(√2))/(16−8−2(√(12)))) =(1/2).((4+(√6)+(√2))/(4−(√(12))))=(1/2).(((4+(√6)+(√2))(4+(√(12))))/4) =(1/8)(16+4(√6)+4(√2)+4(√(12))+6(√2)+2(√(12))) =2+(√(3/2))+(1/(√2))+(√3)+(3/(2(√2)))+((√3)/(2(√2))) =(√4)+((2(√3)+2+2(√6)+3+(√3))/(2(√2)))=(√4)+(√2)+(√3)+(√6)

$$\left.{Method}\:\mathrm{1}\:\right)\:\:{cot}\mathrm{7}.\mathrm{5}°=\frac{{cos}\mathrm{7}.\mathrm{5}°}{{sin}\mathrm{7}.\mathrm{5}}=\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{7}.\mathrm{5}°}{{sin}\mathrm{15}°}=\frac{\mathrm{1}+{cos}\mathrm{15}°}{{sin}\mathrm{15}°}=\frac{\mathrm{2}{cos}\mathrm{15}°+\mathrm{1}+{cos}\mathrm{30}°}{{sin}\mathrm{30}°} \\ $$$$\:\:\:\:\:\:\:\:{cos}\mathrm{15}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\sqrt{\mathrm{2}}}+\mathrm{1}+\sqrt{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${so}\:\:\:\:{cot}\mathrm{7}.\mathrm{5}°=\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{6}}\:\left(\:\:\boldsymbol{{proved}}\right) \\ $$$$ \\ $$$$ \\ $$$${Method}\:\mathrm{2} \\ $$$${cos}\frac{\pi}{\mathrm{24}}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{cos}\frac{\pi}{\mathrm{12}}\right)}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\sqrt{\mathrm{4}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}} \\ $$$${sin}\frac{\pi}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}.\sqrt{\mathrm{4}−\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}} \\ $$$${cot}\frac{\pi}{\mathrm{24}}=\sqrt{\frac{\mathrm{4}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}−\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}}=\frac{\mathrm{4}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{16}−\mathrm{8}−\mathrm{2}\sqrt{\mathrm{12}}}\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{4}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}−\sqrt{\mathrm{12}}}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\mathrm{4}+\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+\sqrt{\mathrm{12}}\right)}{\mathrm{4}} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{16}+\mathrm{4}\sqrt{\mathrm{6}}+\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{12}}+\mathrm{6}\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{12}}\right) \\ $$$$=\mathrm{2}+\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\sqrt{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{4}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{4}}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$$ \\ $$

Commented by bemath last updated on 10/Jul/20

how ((cos 7.5^o )/(sin 7.5^o )) = ((2cos^2 7.5^o )/(sin 7.5^o )) ?

$${how}\:\frac{\mathrm{cos}\:\mathrm{7}.\mathrm{5}^{{o}} }{\mathrm{sin}\:\mathrm{7}.\mathrm{5}^{{o}} }\:=\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{7}.\mathrm{5}^{{o}} }{\mathrm{sin}\:\mathrm{7}.\mathrm{5}^{{o}} }\:? \\ $$

Commented by Dwaipayan Shikari last updated on 10/Jul/20

multiplying by 2cos7.5°

$${multiplying}\:{by}\:\mathrm{2}{cos}\mathrm{7}.\mathrm{5}° \\ $$

Commented by Dwaipayan Shikari last updated on 10/Jul/20

It is sin 15°

$${It}\:{is}\:{sin}\:\mathrm{15}° \\ $$

Commented by bemath last updated on 10/Jul/20

oo sin 15^o . thank

$${oo}\:\mathrm{sin}\:\mathrm{15}^{{o}} .\:{thank} \\ $$

Answered by bobhans last updated on 10/Jul/20

tan 15^o = tan (2×7.5^o ) = ((2tan 7.5^o )/(1−tan^2 (7.5^o ))) (1) tan 15^o =tan (45^o −30^o )=((1−(1/(√3)))/(1+(1/(√3)))) = (((√3)−1)/((√3)+1)) = ((4−2(√3))/2) = 2−(√3) (2) set tan 7.5^o = x (2−(√3))(1−x^2 )= 2x (2−(√3))x^2 +2x−(2−(√3) )=0 x = ((−2+ (√(4+4(2−(√3))^2 )))/(2(2−(√3)))) = ((−1+(√(1+7−4(√3))))/(2−(√3))) = ((−1+(√(8−2(√(12)))))/(2−(√3))) = ((−1+(√6)−(√2))/(2−(√3))) = ((((√6)−(√2)−1)(2+(√3)))/1) = 2(√6)+3(√2)−2(√2)−(√6)−2−(√3)= (√6)+(√2)−2−(√3) = tan 7.5^o

$$\mathrm{tan}\:\mathrm{15}^{{o}} \:=\:\mathrm{tan}\:\left(\mathrm{2}×\mathrm{7}.\mathrm{5}^{{o}} \right)\:=\:\frac{\mathrm{2tan}\:\mathrm{7}.\mathrm{5}^{{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{7}.\mathrm{5}^{{o}} \right)} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{tan}\:\mathrm{15}^{{o}} =\mathrm{tan}\:\left(\mathrm{45}^{{o}} −\mathrm{30}^{{o}} \right)=\frac{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\:= \\ $$$$\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}+\mathrm{1}}\:=\:\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}\: \\ $$$$\left(\mathrm{2}\right)\:{set}\:\mathrm{tan}\:\mathrm{7}.\mathrm{5}^{{o}} \:=\:{x} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\:\mathrm{2}{x} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){x}^{\mathrm{2}} +\mathrm{2}{x}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$${x}\:=\:\frac{−\mathrm{2}+\:\sqrt{\mathrm{4}+\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}\:= \\ $$$$\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{12}}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:= \\ $$$$\frac{−\mathrm{1}+\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:\frac{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{1}}\:= \\ $$$$\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}−\mathrm{2}−\sqrt{\mathrm{3}}= \\ $$$$\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}−\mathrm{2}−\sqrt{\mathrm{3}}\:=\:\mathrm{tan}\:\mathrm{7}.\mathrm{5}^{{o}} \\ $$$$ \\ $$

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