ملتقى الفيزيائيين العرب - عرض مشاركة واحدة

Einstine · #12 05-08-2009, 04:15

$\begin{gathered}\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 8 \hfill \\but:\tan x = \frac{{\sin x}}{{\cos x}},\cot x = \frac{{\cos x}}{{\sin x}},\sec x = \frac{1}{{\cos x}},\csc x = \frac{1}{{\sin x}} \hfill \\\therefore \sin x + \cos x + \frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}} + \frac{1}{{\cos x}} + \frac{1}{{\sin x}} = 8 \hfill \\ \end{gathered}$

$\begin{gathered}\therefore \sin x + \cos x + \frac{{\sin ^2 x + \cos ^2 x}}{{\sin x\cos x}} + \frac{{\sin x + \cos x}}{{\sin x\cos x}} = 8 \hfill \\but:\sin 2x = 2\sin x\cos x \Rightarrow \sin x\cos x = \frac{1}{2}\sin 2x \hfill \\\therefore \sin x + \cos x + \frac{{2\left( {\sin ^2 x + \cos ^2 x} \right)}}{{\sin 2x}} + \frac{{2\left( {\sin x + \cos x} \right)}}{{\sin 2x}} = 8 \hfill \\ \end{gathered}$

$\begin{gathered}\therefore \sin x + \cos x + \frac{2}{{\sin 2x}} + \frac{{2\left( {\cos x + \sin x} \right)}}{{\sin 2x}} = 8 \hfill \\\therefore \sin x + \cos x + \frac{2}{{\sin 2x}}\left( {1 + \cos x + \sin x} \right) = 8 \hfill \\\therefore \sin x + \cos x + \frac{2}{{\sin 2x}}\left( {\cos x + \sin x} \right) = 8 - \frac{2}{{\sin 2x}} \hfill \\ \end{gathered}$

$\begin{gathered}\therefore \left( {\sin x + \cos x} \right)\left( {1 + \frac{2}{{\sin 2x}}} \right) = 8 - \frac{2}{{\sin 2x}} \hfill \\\therefore 1 + \frac{2}{{\sin 2x}} = \frac{{8 - \frac{2}{{\sin 2x}}}}{{\sin x + \cos x}} \Leftrightarrow \left( {1 + \frac{2}{{\sin 2x}}} \right)^2= \frac{{\left( {8 - \frac{2}{{\sin 2x}}} \right)^2 }}{{\left( {\sin x + \cos x} \right)^2 }} \hfill \\ \end{gathered}$

$\therefore 1 + \frac{4}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}} = \frac{{64 - \frac{{32}}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}}}}{{\sin ^2 x + 2\sin x\cos x + \cos ^2 x}}$

$\begin{gathered}\therefore 1 + \frac{4}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}} = \frac{{64 - \frac{{32}}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}}}}{{1 + \sin 2x}} \hfill \\\therefore \left( {1 + \sin 2x} \right)\left( {1 + \frac{4}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}}} \right) = 64 - \frac{{32}}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}} \hfill \\ \end{gathered}$

$\begin{gathered}\therefore 1 + \sin 2x + \frac{4}{{\sin 2x}} + 4 + \frac{4}{{\sin ^2 2x}} + \frac{4}{{\sin 2x}} = 64 - \frac{{32}}{{\sin 2x}} + \frac{4}{{\sin ^2 2x}} \hfill \\\therefore \sin 2x + \frac{{40}}{{\sin 2x}} - 59 = 0 \hfill \\\therefore \sin ^2 2x - 59\sin 2x + 40 = 0 \hfill \\ \end{gathered}$

$\begin{gathered}\therefore \sin ^2 2x - 59\sin 2x + 40 = 0 \hfill \\let:\sin 2x = u \hfill \\\therefore u^2- 59u + 40 = 0 \hfill \\ \end{gathered}$

$\begin{gathered}if:ax^2+ bx + c = 0 \hfill \\\therefore x_1= \frac{{ - b + \sqrt {b^2- 4ac} }}{{2a}},x_2= \frac{{ - b - \sqrt {b^2- 4ac} }}{{2a}} \hfill \\ \end{gathered}$

$\begin{gathered}\therefore u_1= \frac{{59 + \sqrt {3481 - 160} }}{2} = \frac{{59 + \sqrt {3321} }}{2} = \frac{{59 + 9\sqrt {41} }}{2} \approx 58.3141 \hfill \\,u_2= \frac{{59 - \sqrt {3481 - 160} }}{2} = \frac{{59 - \sqrt {3321} }}{2} = \frac{{59 - 9\sqrt {41} }}{2} \approx 0.6859 \hfill \\\therefore \sin 2x = 58.3141\text{ }or\text{ }\sin 2x = 0.6859 \hfill \\ \end{gathered}$