Identitate trigonometrica

Mateescu Constantin · Post by **Mateescu Constantin** » Fri Jul 31, 2009 3:59 pm

Aratati ca \( \frac {1}{4\cos^2 50^{\circ}} + 4\sin^2 10^{\circ} = 4\sin^2 20^{\circ} + \frac {1}{4\sin^2 100^{\circ}} \).

Marius Mainea · Post by **Marius Mainea** » Fri Nov 13, 2009 11:55 pm

Inegalitatea este echivalenta cu

\( \frac{1}{4}\frac{\sin 40^\circ\sin60^\circ}{\cos^250^\circ\cos10^\circ}=4\sin10^\circ\sin30^\circ \)

sau

\( \frac{\sqrt{3}}{8}=\sin40\sin80\sin20 \)

care este adevarat daca se transforma in suma.

DrAGos Calinescu · Post by **DrAGos Calinescu** » Sat Nov 14, 2009 11:59 am

\( \sin 3t=4\sin t \sin (60 - t)\sin(60 + t) \)
\( \Longrightarrow\ \sin 40\sin80\sin20=\frac{\sin 60}{4}=\frac{\sqrt{3}}{8} \)