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Bonsoir. Comment lineariser sin(π/30)*sin(7π/30) ? merci d'avance.

Répondre :

[tex]sin(\frac{\pi}{30})*sin(\frac{7\pi}{30})= \frac{e^{i\frac{\pi}{30}}-e^{-i\frac{\pi}{30}}}{2i}*\frac{e^{i\frac{7\pi}{30}}-e^{-i\frac{7\pi}{30}}}{2i}=\frac{(e^{i\frac{\pi}{30}}-e^{-i\frac{\pi}{30}})(e^{i\frac{7\pi}{30}}-e^{-i\frac{7\pi}{30}})}{(2i)^2}[/tex][tex]=\frac{e^{i\frac{\pi}{30}+\frac{7\pi}{30}}-e^{i\frac{\pi}{30}-\frac{7\pi}{30}}-e^{-i\frac{\pi}{30}+\frac{7\pi}{30}}+e^{-i\frac{\pi}{30}-\frac{7\pi}{30}}}{-4}=-\frac{e^{i\frac{8\pi}{30}}-e^{-i\frac{6\pi}{30}}-e^{i\frac{6\pi}{30}}+e^{-i\frac{8\pi}{30}}}{4}[/tex][tex]=\frac{(e^{i\frac{6\pi}{30}}+e^{-i\frac{6\pi}{30}})-(e^{i\frac{8\pi}{30}}+e^{-i\frac{8\pi}{30}})}{4}=\frac{2cos(\frac{6\pi}{30})-2cos(\frac{8\pi}{30})}{4}[/tex][tex]= \frac{cos(\frac{6\pi}{30})-cos(\frac{8\pi}{30})}{2}=\frac{1}{2}(cos(\frac{6\pi}{30})-cos(\frac{8\pi}{30}))=\frac{1}{2}(cos(\frac{\pi}{5})-cos(\frac{4\pi}{15}))[/tex]