O relatie trigonometrica particulara
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
O relatie trigonometrica particulara
Sa se arate ca \( \tan 10^{\circ}=\tan 20^{\circ}\cdot\tan 30^{\circ}\cdot\tan 40^{\circ} \) .
Iata o demonstratie cat se poate de "ne-spectaculoasa" 
.
Identitatea este echivalenta succesiv cu:
\( sin10^{\circ}cos20^{\circ}cos30^{\circ}cos40^{\circ}=cos10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\ \)
\( 8cos10^{\circ}sin10^{\circ}cos20^{\circ}cos30^{\circ}cos40^{\circ}=8cos^{2}10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\
sin80^{\circ}cos30^{\circ}=8cos^{2}10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\ \)
\( cos30^{\circ}=8sin20^{\circ}sin30^{\circ}sin40^{\circ}sin80^{\circ}\Leftrightarrow \\ \)
\( sin20^{\circ}sin40^{\circ}sin60^{\circ}sin80^{\circ}=\frac{3}{16}. \) \( (*) \)
Identitatea \( (*) \) este cunoscuta. Sa o demonstram:
\( sin20^{\circ}sin40^{\circ}sin60^{\circ}sin80^{\circ}=\frac{(cos20^{\circ}-cos60^{\circ})(cos20^{\circ}-cos140^{\circ})}{4}=\\ \frac{(cos20^{\circ}-\frac{1}{2})(cos20^{\circ}+2cos^{2}20^{\circ}-1)}{4}=\frac{4cos^{3}20^{\circ}-3cos20^{\circ}+1}{8}=\\ \frac{cos60^{\circ}+1}{8}=\frac{3}{16}. \)
. Identitatea este echivalenta succesiv cu:
\( sin10^{\circ}cos20^{\circ}cos30^{\circ}cos40^{\circ}=cos10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\ \)
\( 8cos10^{\circ}sin10^{\circ}cos20^{\circ}cos30^{\circ}cos40^{\circ}=8cos^{2}10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\
sin80^{\circ}cos30^{\circ}=8cos^{2}10^{\circ}sin20^{\circ}sin30^{\circ}sin40^{\circ}\Leftrightarrow \\ \)
\( cos30^{\circ}=8sin20^{\circ}sin30^{\circ}sin40^{\circ}sin80^{\circ}\Leftrightarrow \\ \)
\( sin20^{\circ}sin40^{\circ}sin60^{\circ}sin80^{\circ}=\frac{3}{16}. \) \( (*) \)
Identitatea \( (*) \) este cunoscuta. Sa o demonstram:
\( sin20^{\circ}sin40^{\circ}sin60^{\circ}sin80^{\circ}=\frac{(cos20^{\circ}-cos60^{\circ})(cos20^{\circ}-cos140^{\circ})}{4}=\\ \frac{(cos20^{\circ}-\frac{1}{2})(cos20^{\circ}+2cos^{2}20^{\circ}-1)}{4}=\frac{4cos^{3}20^{\circ}-3cos20^{\circ}+1}{8}=\\ \frac{cos60^{\circ}+1}{8}=\frac{3}{16}. \)