Question Number 101116 by Jamshidbek2311 last updated on 30/Jun/20

Answered by 1549442205 last updated on 07/Jul/20

We prove that tan((3𝛑)/(11))+4sin ((2𝛑)/(11))=(√(11))  ⇔sin((3π)/(11))+4sin((2π)/(11))cos((3π)/(11))=(√(11))cos((3π)/(11))  ⇔3sin(π/(11))−4sin^3 (π/(11))+4(2sin(π/(11))cos(π/(11))− (√(11)) )( 4cos^3 (π/(11))−3cos(π/(11)))=0  ⇔sin(π/(11))[3−4(1−cos^2 (π/(11)))]+4(2sin(π/(11))cos(π/(11))− (√(11)) )( 4cos^3 (π/(11))−3cos(π/(11)))=0  ⇔sin(π/(11))(4cos^2 (π/(11))−1)+4(2sin(π/(11))cos(π/(11))− (√(11)) )( 4cos^3 (π/(11))−3cos(π/(11)))=0  Putting   cos(π/(11))=x ⇒sin(π/(11))=(√(1−x^2 )) we get  (√(1−x^2 ))(4x^2 −1)+4(2x(√(1−x^2 ))−(√(11)))(4x^3 −3x)=0  ⇔(√(1−x^2 ))(32x^4 −20x^2 −1)=(√(11))(4x^3 −3x)  Square two sides of above we get  (1−x^2 )(1024x^8 −1280x^6 +336x^4 +40x^2 +1)=11(16x^6 −24x^4 +9x^2 )  ⇔(−1024x^(10) +2304x^8 −1616x^6 +296x^4 +39x^2 +1=176x^6 −264x^4 +99x^2   ⇔1024x^(10) −2304x^8 +1792x^6 −560x^4 +60x^2 −1=0  =(32x^5 −16x^4 −32x^3 +12x^2 +6x−1)(32x^5 +16x^4 −32x^3 −12x^2 +6x+1)=0  ⇔(32x^5 −16x^4 −32x^3 +12x^2 +6x−1)=0 (∗)  (because (32x^5 +16x^4 −32x^3 −12x^2 +6x+1≠0 for x=(π/(11)))  Now we need prove the equality is true.  Indeed,  Applying Mauvra′s formular we have  (cos(π/(11))+isin(π/(11)))^5 =cos((5π)/(11))+isin((5π)/(11))  ⇔cos^5 (π/(11))+5icos^4 (π/(11))sin(π/(11))+10i^2 cos^3 (π/(11))sin^2 (π/(11))+10cos^2 .i^3 sin^3 (π/(11))+5cos(π/(11)).i^4 sin^4 (π/(11))+sin^5 (π/(11))=cos((5π)/(11))+isin((5π)/(11))   ⇒cos((5𝛑)/(11))=cos^5 (π/(11))−10cos^3 (π/(11))sin^2 (π/(11))+5cos(π/(11))sin^4 (π/(11))  =cos^5 (π/(11))−10cos^3 (π/(11))(1−cos^2 (π/(11)))+5cos(π/(11))(1−cos^2 (π/(11)))^2   =16cos^5 (𝛑/(11))−20cos^3 (𝛑/(11))+5cos(𝛑/(11))(1)  (cos(π/(11))+isin(π/(11)))^6 =cos((6π)/(11))+isin((6π)/(11))=−cos((5π)/(11))+isin((5π)/(11))  cos^6 (π/(11))+6cos^5 (π/(11)).isin(π/(11))+15cos^4 (π/(11)).i^2 sin^2 (π/(11))+20cos^3 (π/(11)).i^3 sin^3 (π/(11))+15cos^2 (π/(11)).i^4 sin^4 (π/(11))  +5cos(π/(11)).i^5 sin^5 (π/(11))+sin^6 (π/(11))=−cos((5π)/(11))+isin((5π)/(11))  ⇔x^6 −15x^4 y^2 +15x^2 y^4 +y^6 +i.(6x^5 y−15x^4 y^2 −20x^3 y^3 )=−cos((5𝛑)/(11))+isin((5𝛑)/(11))(y=sin(𝛑/(11)))(2)  From (1) (2) we get  x^6 −15x^4 y^2 +15x^2 y^4 +y^6 =−16x^5 +20x^3 −5x   ⇔x^6 −15x^4 (1−x^2 )+15x^2 (1−x^2 )^2 +(1−x^2 )^3 =−16x^5 +20x^3 −5x  ⇔32x^6 −48x^4 +18x^2 −1=−16x^5 +20x^3 −5x  ⇔32x^6 +16x^6 −48x^4 −20x^3 +18x^2 +5x−1=0  ⇔(x+1)(16x^5 −16x^4 −32x^3 +12x^2 +6x−1)=0  ⇔16x^5 −16x^4 −32x^3 +12x^2 +6x−1=0  Thus,the equality (∗)proved   Therefore,tan((3𝛑)/(11))+4sin((2𝛑)/(11))=(√(11 )) (q.e.d)