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Data publicarii: 06 Noiembrie, 2014

EXERCITIUL 13

Suport teoretic:

Inegalitati,sume,integrale definite,primitive directe,progresii,ecuatii trigonometrice.

Enunt:

Sa se rezolve ecuatia:

\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\frac{1-sin^nx}{2sinx},\;unde\;{x}\in{(0,\frac{\pi}{2})},\;{t}>{0},\;{n}\in{\mathbb{N^{*}}}.\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\frac{1-sin^nx}{2sinx},\;unde\;{x}\in{(0,\frac{\pi}{2})},\;{t}>{0},\;{n}\in{\mathbb{N^{*}}}.

Raspuns:

x = π/6.

Rezolvare:

Sa calculam membrul intai al inegalitatii:

\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\sum_{k=0}^{k=n-1}{\big((k+1)\cdot{\frac{t^{k+1}}{k+1}}|_0^{sinx}\big)}=\sum_{k=0}^{k=n-1}{\big(t^{k+1}\big|_0^{sinx}\big)}=\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\sum_{k=0}^{k=n-1}{\big((k+1)\cdot{\frac{t^{k+1}}{k+1}}|_0^{sinx}\big)}=\sum_{k=0}^{k=n-1}{\big(t^{k+1}\big|_0^{sinx}\big)}= =\sum_{k=0}^{k=n-1}{\big({sin}^{k+1}x-sin^{k+1}0\big)}=sinx+sin^2x+sin^3x+\cdots+sin^nx==\sum_{k=0}^{k=n-1}{\big({sin}^{k+1}x-sin^{k+1}0\big)}=sinx+sin^2x+sin^3x+\cdots+sin^nx=

=(sinx)\cdot{\frac{1-sin^nx}{1-sinx}}.=(sinx)\cdot{\frac{1-sin^nx}{1-sinx}}.

Inegalitatea de demonstrat devine, succesiv:

{(sinx)\cdot{\frac{1-sin^nx}{1-sinx}}=\frac{1-sin^nx}{2sinx}}\Leftrightarrow{2sin^2x=1-sinx}\Leftrightarrow{2sin^2x+sinx-1=0}\Leftrightarrow{(sinx)\cdot{\frac{1-sin^nx}{1-sinx}}=\frac{1-sin^nx}{2sinx}}\Leftrightarrow{2sin^2x=1-sinx}\Leftrightarrow{2sin^2x+sinx-1=0}\Leftrightarrow

\Leftrightarrow{sinx=\frac{1}{2},\;sau\;sinx=-1}.\Leftrightarrow{sinx=\frac{1}{2},\;sau\;sinx=-1}.

Convine sinx = 1/2, deci x = π/6.

Postat în: SUME-liceu

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