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Question Number 189256 by BaliramKumar last updated on 13/Mar/23
Prove that  sin10° = (1/2)(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−...........∞))))))))))))))))))))))))))
$${Prove}\:{that} \\ $$$$\mathrm{sin10}°\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−………..\infty}}}}}}}}}}}}} \\ $$
Answered by Frix last updated on 14/Mar/23
sin 10° =sin (π/(18))  First let′s find x=(√(2−(√(2+(√(2+x))))))  3 times squaring and transforming leads to  x^8 −8x^6 +20x^4 −16x^2 −x+2=0  Fortunately x=−1∧x=2 are solutions  (but they′re both false)  Now we′ve got  x^6 +x^5 −5x^4 −3x^3 +7x^2 +x−1=0  (x^3 −3x+1)(x^3 +x^2 −2x−1)=0  The first is:  (x+2cos (π/9))(x−2sin (π/(18)))(x−2cos ((2π)/9))=0  (The second is irrelevant.)  ⇒  2sin (π/(18)) =(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−...))))))))))))))
$$\mathrm{sin}\:\mathrm{10}°\:=\mathrm{sin}\:\frac{\pi}{\mathrm{18}} \\ $$$$\mathrm{First}\:\mathrm{let}'\mathrm{s}\:\mathrm{find}\:{x}=\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{x}}}} \\ $$$$\mathrm{3}\:\mathrm{times}\:\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{8}} −\mathrm{8}{x}^{\mathrm{6}} +\mathrm{20}{x}^{\mathrm{4}} −\mathrm{16}{x}^{\mathrm{2}} −{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{Fortunately}\:{x}=−\mathrm{1}\wedge{x}=\mathrm{2}\:\mathrm{are}\:\mathrm{solutions} \\ $$$$\left(\mathrm{but}\:\mathrm{they}'\mathrm{re}\:\mathrm{both}\:\mathrm{false}\right) \\ $$$$\mathrm{Now}\:\mathrm{we}'\mathrm{ve}\:\mathrm{got} \\ $$$${x}^{\mathrm{6}} +{x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{is}: \\ $$$$\left({x}+\mathrm{2cos}\:\frac{\pi}{\mathrm{9}}\right)\left({x}−\mathrm{2sin}\:\frac{\pi}{\mathrm{18}}\right)\left({x}−\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right)=\mathrm{0} \\ $$$$\left(\mathrm{The}\:\mathrm{second}\:\mathrm{is}\:\mathrm{irrelevant}.\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{2sin}\:\frac{\pi}{\mathrm{18}}\:=\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−…}}}}}}} \\ $$
Commented by BaliramKumar last updated on 14/Mar/23
thanks  sir
$${thanks}\:\:{sir} \\ $$

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