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Question Number 196008 by universe last updated on 15/Aug/23

Commented by universe last updated on 15/Aug/23

prove that

provethat

Commented by York12 last updated on 15/Aug/23

what is the source of those

whatisthesourceofthose

Answered by MM42 last updated on 17/Aug/23

According to the “ de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) (sinα)− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 + (((2n+1)),(( 5)) )(cosα)^(2n−3) (sinα)^5 +... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) )(tanα)^2 + (((2n+1)),(( 5)) )(tanα)^4 − (((2n+1)),(( 7)) )(tanα)^6 +...] for α_k =((kπ)/(2n+1)) ; ∀ 1≤k≤n ⇒ sin(2n+1)α=0 ⇒ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) )(tanα_k )^2 + (((2n+1)),(( 5)) )(tanα_k )^4 −....=0 ⇒x_k =(tanα_k )^2 ; 1≤k≤n the roots of equation are below x^n − (((2n+1)),((2n−1)) )x^(n−1) + (((2n+1)),((2n−3)) )x^(n−2) −....=0 the sume of the roots “ s= (((2n+1)),((2n−1)) ) ” ⇒Σ_(k=1) ^n (tan((kπ)/(2n+1)))^2 =(((2n+1)!)/((2n−1)!))=n(2n+1) ✓ the second part is similary proved

Accordingtothe‘‘demoivre″sin(2n+1)α=(2n+11)(cosα)2n(sinα)−(2n+13)(cosα)2n−2(sinα)3+(2n+15)(cosα)2n−3(sinα)5+...=(cosα)2n(sinα)[(2n+11)−(2n+13)(tanα)2+(2n+15)(tanα)4−(2n+17)(tanα)6+...]forαk=kπ2n+1;∀1⩽k⩽n⇒sin(2n+1)α=0⇒(2n+11)−(2n+13)(tanαk)2+(2n+15)(tanαk)4−....=0⇒xk=(tanαk)2;1⩽k⩽ntherootsofequationarebelowxn−(2n+12n−1)xn−1+(2n+12n−3)xn−2−....=0thesumeoftheroots‘‘s=(2n+12n−1)″⇒∑nk=1(tankπ2n+1)2=(2n+1)!(2n−1)!=n(2n+1)✓thesecondpartissimilaryproved

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