How to prove that:tan⁡(3π11)+4sin⁡(2π11)=11

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mtakadamu9i5 · Accepted Answer

Since tan⁡3π11+4sin⁡2π11&amp;gt;0, it&#039;s enough to prove that(sin⁡3π11+4sin⁡2π11cos⁡3π11)2=11cos23π11or(sin⁡3π11+2sin⁡5π11−2sin⁡π11)2=11cos23π11or1−cos⁡6π11+4−4cos⁡10π11+4−4cos⁡2π11+4cos⁡2π11−4cos⁡8π11−4cos⁡2π11+4cos⁡4π11−8cos⁡4π11−8cos⁡4π11+8cos⁡6π11=11+11cos⁡6π11or∑k=15cos⁡2kπ11=−12or∑k=152sin⁡π11cos⁡2kπ11=−sin⁡π11or∑k=15(sin⁡(2k+1)π11−sin⁡(2k−1)π11)=−sin⁡π11orsin⁡11π11−sin⁡π11=−sin⁡π11Done!

bedevijuo3e · Accepted Answer

x=tan⁡(3π11)+4sin⁡(2π11)For simpliying equation, I usedu=π11and11u=πtransformations,Hence,x=tan⁡3u+4sin⁡2u2cos⁡3u⋅x=2cos⁡3u⋅tan⁡3u+8cos⁡3u⋅sin⁡2uHence,2cos⁡3u⋅x=2sin⁡3u+8cos⁡3u⋅sin⁡2uAfter squaring both sides,(2cos⁡3u⋅x)2=(2sin⁡3u+8cos⁡3u⋅sin⁡2u)24(cos⁡3u)2⋅x2=4(sin⁡3u)2+32sin⁡3u⋅cos⁡3u⋅sin⁡2u+64(cos⁡3u)2⋅(sin⁡2u)2=2⋅(1−cos⁡6u)+16sin⁡6u⋅sin⁡2u+16⋅(1+cos⁡6u)⋅(1−cos⁡4u)=2−2cos⁡6u+8cos⁡4u−8cos⁡8u+16+16cos⁡6u−16cos⁡4u−16cos⁡6u⋅cos⁡4u=18+14cos⁡6u−8cos⁡4u−8cos⁡8u−8⋅(cos⁡10u+cos⁡2u)=18+14cos⁡6u−8cos⁡4u−8co8u−8cos⁡10u−8cos⁡2uAfter multiplying both sides with sin⁡u,

How to prove that: \tan(3\pi/11)+4\sin(2\pi/11)=\sqrt{11}

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