what-is-tan-3pi-11-4sin-2pi-11-

Question Number 81286 by jagoll last updated on 11/Feb/20

w h a t i s

\tan \frac{3 π}{11} + 4 \sin \frac{2 π}{11} ?

Answered by MJS last updated on 11/Feb/20

a strange way to solve this

\sin α = \frac{e^{i α} - e^{- i α}}{2 i}; \cos α = \frac{e^{i α} + e^{- i α}}{2}

let e^{i \frac{3 π}{11}} = u^{\frac{3}{2}} \land e^{i \frac{2 π}{11}} = u

\frac{\frac{u^{\frac{3}{2}} - u^{- \frac{3}{2}}}{2 i}}{\frac{u^{\frac{3}{2}} + u^{- \frac{3}{2}}}{2}} + 4 \frac{u - u^{- 1}}{2 i} = t > 0

- i \frac{2 u^{5} + u^{4} - 2 u^{3} + 2 u^{2} - u - 2}{u^{4} + u} = t

squaring this \Rightarrow

- \frac{{(u - 1)}^{2} {(2 u^{4} + 3 u^{3} + u^{2} + 3 u + 2)}^{2}}{u^{2} {(u + 1)}^{2} {(u^{2} - u + 1)}^{2}} = t^{2}

\Rightarrow

4 u^{10} + 4 u^{9} + (t^{2} - 7) u^{8} + 4 u^{7} + 4 u^{6} + 2 (t^{2} - 9) u^{5} + 4 u^{4} + 4 u^{3} + (t^{2} - 7) u^{2} + 4 u + 4 = 0

now if t^{2} - 7 = 2 (t^{2} - 9) = 4 this would be nice

\Rightarrow t = \pm \sqrt{11} but we know that t > 0 \Rightarrow t = \sqrt{11}

\Rightarrow

u^{10} + u^{9} + u^{8} + u^{7} + u^{6} + u^{5} + u^{4} + u^{3} + u^{2} + u + 1 = 0

multiplicate with (u - 1)

\Rightarrow

u^{11} - 1 = 0 \Rightarrow u = 1 \lor u = e^{i \frac{n π}{11}} with n \in N \land 1 ⩽ n ⩽ 10

but u = e^{i \frac{2 π}{11}} which indeed is a solution and

u^{\frac{3}{2}} = e^{i \frac{3 π}{11}} which is also a solution

\Rightarrow

t = \sqrt{11}

Commented by jagoll last updated on 11/Feb/20

v i a c o m p l e x a n a l y s i s s i r

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