Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$${thanks}\:\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com