Θέμα 16

ghan · #1

Αν $\displaystyle\alpha ,\phi \in \left( 0,\frac{\pi }{2} \right)$ , τότε $\displaystyle\phi <\int_{0}^{\phi }{\frac{dx}{\sqrt{1-{{\sin }^{2}}\alpha \cdot {{\sin }^{2}}x}}}<\frac{\phi }{\sqrt{1-{{\sin }^{2}}\alpha \cdot {{\sin }^{2}}\phi }}$

mathxl · #2

Για την μία φορά $\displaystyle{0 \le {\sin ^2}a \le 1 \Rightarrow 1 \ge 1 - {\sin ^2}\alpha \cdot{\sin ^2}x \ge {\cos ^2}x \Rightarrow 1 \le \frac{1}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}x} }} \le \frac{1}{{\cos x}} \Rightarrow }$
$\displaystyle{\phi < \int\limits_0^\phi {\frac{{dx}}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}x} }}} < \int\limits_0^\phi {\frac{{dx}}{{\cos x}}} }$

mathxl · #3

Η άλλη φορά

$\displaystyle{i{n^2}x \le {\sin ^2}\phi \Rightarrow \sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}x} \ge \sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}\phi } \Rightarrow \frac{1}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}x} }} \le \frac{1}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}\phi } }} \Rightarrow }$
$\displaystyle{\int\limits_0^\phi {\frac{{dx}}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}x} }}} < \int\limits_0^\phi {\frac{{dx}}{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}\phi } }}} = \frac{\phi }{{\sqrt {1 - {{\sin }^2}\alpha \cdot{{\sin }^2}\phi } }}}$

Θέμα 16

Θέμα 16

Re: Θέμα 16

Re: Θέμα 16

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