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let-Q-1-tan-3pi-8-tan-pi-10-1-tan-pi-8-tan-pi-10-prove-that-Q-1-Q-1-7-3-5-85-38-5-

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Question Number 72332 by aliesam last updated on 27/Oct/19
let Q=((1+tan(((3π)/8)) . tan((π/(10))))/(1−tan((π/8)).tan((π/(10))))) prove that ((Q−1)/(Q+1))=(√(7−3(√5)−(√(85−38(√5)))))
$${let}\:{Q}=\frac{\mathrm{1}+{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:.\:{tan}\left(\frac{\pi}{\mathrm{10}}\right)}{\mathrm{1}−{tan}\left(\frac{\pi}{\mathrm{8}}\right).{tan}\left(\frac{\pi}{\mathrm{10}}\right)} \\ $$$$ \\ $$$${prove}\:{that} \\ $$$$\: \\ $$$$\frac{{Q}−\mathrm{1}}{{Q}+\mathrm{1}}=\sqrt{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}−\sqrt{\mathrm{85}−\mathrm{38}\sqrt{\mathrm{5}}}} \\ $$$$ \\ $$
Answered by mind is power last updated on 27/Oct/19
((Q−1)/(Q+1))=((tan((π/(10)))(tan(((3π)/8))+tan((π/8))))/(2+tan((π/(10)))(tan(((3π)/8))−tan((π/8)))) ((sin(a+_− b))/(cos(a)cos(b)))=tg(a)^ +_− tg(b) cos(a)cos(b)=(1/2)(cos(a−b)+cos(a+b)) ⇒tg(a)+tg(b)=((2sin(a+b))/(cos(a+b)+cos(a−b))) ⇒tan(((3π)/8))+tan((π/8))=((2sin((π/2)))/(cos((π/4))))=2(√2) tan(((3π)/8))−tan((π/8))=((2sin((π/4)))/(cos((π/4))))=2 ((Q−1)/(Q+1))=((2(√2)tan((π/(10))))/(2+2tan((π/(10)))))=(((√2)tan((π/(10))))/(1+tan((π/(10)))))=(√((2tg^2 ((π/(10))))/((1+tg((π/(10))))^2 ))) ((tg^2 (x))/((1+tg(x))^2 ))=f(x)=((2sin^2 (x))/((1+sin(2x))))=((1+cos(2x))/((1+sin(2x))) x=(π/(10)) ⇒f((π/(10)))=((1+cos((π/5)))/((1+sin((π/5))))) cos(((2π)/5))=((−1+(√5))/4)=2cos^2 ((π/5))−1=−2sin^2 ((π/5))+1 too bee continued
$$\frac{\mathrm{Q}−\mathrm{1}}{\mathrm{Q}+\mathrm{1}}=\frac{\mathrm{tan}\left(\frac{\pi}{\mathrm{10}}\right)\left(\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}\right)\right)}{\mathrm{2}+\mathrm{tan}\left(\frac{\pi}{\mathrm{10}}\right)\left(\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}\right)\right.} \\ $$$$\frac{\mathrm{sin}\left(\mathrm{a}\underset{−} {+}\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right)}=\mathrm{tg}\left(\mathrm{a}\overset{} {\right)}\underset{−} {+}\mathrm{tg}\left(\mathrm{b}\right) \\ $$$$\mathrm{cos}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)\right) \\ $$$$\Rightarrow\mathrm{tg}\left(\mathrm{a}\right)+\mathrm{tg}\left(\mathrm{b}\right)=\frac{\mathrm{2sin}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)} \\ $$$$\Rightarrow\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{2sin}\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{2sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)}=\mathrm{2} \\ $$$$\frac{\mathrm{Q}−\mathrm{1}}{\mathrm{Q}+\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{10}}\right)}{\mathrm{2}+\mathrm{2tan}\left(\frac{\pi}{\mathrm{10}}\right)}=\frac{\sqrt{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{10}}\right)}{\mathrm{1}+\mathrm{tan}\left(\frac{\pi}{\mathrm{10}}\right)}=\sqrt{\frac{\mathrm{2tg}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{10}}\right)}{\left(\mathrm{1}+\mathrm{tg}\left(\frac{\pi}{\mathrm{10}}\right)\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{tg}\left(\mathrm{x}\right)\right)^{\mathrm{2}} }=\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{sin}\left(\mathrm{2x}\right)\right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\left(\mathrm{1}+\mathrm{sin}\left(\mathrm{2x}\right)\right.} \\ $$$$\mathrm{x}=\frac{\pi}{\mathrm{10}}\:\Rightarrow\mathrm{f}\left(\frac{\pi}{\mathrm{10}}\right)=\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)}{\left(\mathrm{1}+\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)\right)} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}=\mathrm{2cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1}=−\mathrm{2sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)+\mathrm{1} \\ $$$$\mathrm{too}\:\mathrm{bee}\:\mathrm{continued} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aliesam last updated on 27/Oct/19
perfect sir
$${perfect}\:{sir}\: \\ $$

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