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> Pomoc Sa Trigonometrijom

username
post Sep 6 2008, 04:19 PM
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Group: Članovi
Joined: 29-September 07
Member No.: 684
Status: Bivši učenik MGa



mozete li da mi pomognete oko ovog zadatka:

sin^3a+cos^3a
--------------------
sin^3a-cos^3a

izracunati vrednost ako je tga=2
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vriskica
post Sep 6 2008, 05:00 PM
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Group: Članovi
Joined: 23-August 08
Member No.: 1.085
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probaj ovako..

sin3a+cos3a=(sina+cosa)(sin2a-sina*cosa+cos2a)
sin3a-cos3a=(sina-cosa)(sin2a+sina*cosa+cos2a)
sin2a+cos2a=1
(sina+cosa)(sin2a-sina*cosa+cos2a) / sina-cosa)(sin2a+sina*cosa+cos2a)=(sina+cosa)*-sina*cosa / =(sina-cosa)*sina*cosa =-(sina+cosa) / (sina-cosa)

sad i brojnik i nazivnik podjelis sa cosa

i dobit ces
-tg-1 / tg-1=-2-1/2-1=-3
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Posts in this topic
username   Pomoc Sa Trigonometrijom   Sep 6 2008, 04:19 PM
vriskica   RE: Pomoc Sa Trigonometrijom   Sep 6 2008, 05:00 PM
username   RE: Pomoc Sa Trigonometrijom   Sep 6 2008, 05:12 PM
vriskica   RE: Pomoc Sa Trigonometrijom   Sep 6 2008, 05:13 PM


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