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Question-179226

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Question Number 179226 by mnjuly1970 last updated on 26/Oct/22
Commented by mnjuly1970 last updated on 26/Oct/22
prove ⇑⇑
$$\:\:\:{prove}\:\Uparrow\Uparrow \\ $$
Answered by Acem last updated on 27/Oct/22
S_1 = (√2) (√(3+4 cos^3 ((2π)/7))) = (√2) (√(3+cos 2π +3 cos ((2π)/7))) = (√2) (√(4+ 3 cos ((2π)/7))) You can continue and give me a hint (:
$$\:{S}_{\mathrm{1}} =\:\sqrt{\mathrm{2}}\:\sqrt{\mathrm{3}+\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:\frac{\mathrm{2}\pi}{\mathrm{7}}}\:=\:\sqrt{\mathrm{2}}\:\sqrt{\mathrm{3}+\mathrm{cos}\:\mathrm{2}\pi\:+\mathrm{3}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}} \\ $$$$\:\:\:\:\:\:=\:\sqrt{\mathrm{2}}\:\sqrt{\mathrm{4}+\:\mathrm{3}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}}\: \\ $$$$\:{You}\:{can}\:{continue}\:{and}\:{give}\:{me}\:{a}\:{hint}\:\left(:\right. \\ $$
Commented by Frix last updated on 27/Oct/22
4cos^3 ((2π)/7) ≠ cos 2π +3cos ((2π)/7) 4cos^3 ((2π)/7) =cos ((6π)/7) +3cos ((2π)/7)
$$\mathrm{4cos}^{\mathrm{3}} \:\frac{\mathrm{2}\pi}{\mathrm{7}}\:\neq\:\mathrm{cos}\:\mathrm{2}\pi\:+\mathrm{3cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$\mathrm{4cos}^{\mathrm{3}} \:\frac{\mathrm{2}\pi}{\mathrm{7}}\:=\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{7}}\:+\mathrm{3cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$
Answered by Frix last updated on 27/Oct/22
squaring and transforming with lots of basic trigonometric formulas leads to sin ((5π)/(14)) −sin ((3π)/(14)) +sin (π/(14)) =(1/2) which we have to prove
$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{with}\:\mathrm{lots}\:\mathrm{of} \\ $$$$\mathrm{basic}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{which}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove} \\ $$
Commented by Frix last updated on 28/Oct/22
I found an unexpected proof: sin ((5π)/(14)) −sin ((3π)/(14)) +sin (π/(14)) =r (sin ((5π)/(14)) −sin ((3π)/(14)) +sin (π/(14)))^2 =r^2 sin^2 ((5π)/(14)) +sin^2 ((3π)/(14)) +sin^2 (π/(14)) −2(sin ((5π)/(14)) sin ((3π)/(14)) −sin ((5π)/(14)) sin (π/(14)) +sin ((3π)/(14)) sin (π/(14)))=r^2 ((1+sin ((3π)/(14)))/2)+((1−sin (π/(14)))/2)+((1−sin ((5π)/(14)))/2)−2(((sin ((5π)/(14)) +sin (π/(14)))/2)−((sin ((3π)/(14)) −sin (π/(14)))/2)+((sin ((5π)/(14)) −sin ((3π)/(14)))/2))=r^2 −((5(sin ((5π)/(14)) −sin ((3π)/(14)) +sin (π/(14))))/2)+(3/2)=r^2 −((5r)/2)+(3/2)=r^2 r^2 +((5r)/2)−(3/2)=0 r=−3 obviously not possible ⇒ r=(1/2) q.e.d
$$\mathrm{I}\:\mathrm{found}\:\mathrm{an}\:\mathrm{unexpected}\:\mathrm{proof}: \\ $$$$\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:={r} \\ $$$$\left(\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{5}\pi}{\mathrm{14}}\:+\mathrm{sin}^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{14}}\:−\mathrm{2}\left(\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)={r}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}}{\mathrm{2}}+\frac{\mathrm{1}−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}}{\mathrm{2}}+\frac{\mathrm{1}−\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}}{\mathrm{2}}−\mathrm{2}\left(\frac{\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}}{\mathrm{2}}−\frac{\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}}{\mathrm{2}}+\frac{\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}}{\mathrm{2}}\right)={r}^{\mathrm{2}} \\ $$$$−\frac{\mathrm{5}\left(\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}={r}^{\mathrm{2}} \\ $$$$−\frac{\mathrm{5}{r}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}={r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{5}{r}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$${r}=−\mathrm{3}\:\mathrm{obviously}\:\mathrm{not}\:\mathrm{possible} \\ $$$$\Rightarrow\:{r}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{q}.\mathrm{e}.\mathrm{d} \\ $$

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