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XxWiLLxX
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Pe: 2014-03-11T19:12:30+02:00 In: In:

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Va rog frum sa ma ajutati … Mi sa spus ca aceasta prob va fi la testul de maine si nuj cum se face.. :
Aratati ca [sin^2(3x)]/(sin^2x) – [cos^2(3x)]/cos^2x = 8cos2x , sinx nui egal cu 0 si cosx nui egal cu 0
Va rog ajutatima … Multumesc anticipat 😀

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  1. maestru (V)

    \it{\bl \Large sin3x = 3sin x-4sin^3x = sinx (3-4sin^2x) (*)\\\;\\cos\3x = -3cos x + 4cos^3x = cosx (-3+4cos^2x) (**)\\\;\\ \\\;\\\frac{sin^2 3x}{sin^2 x} - \frac{cos^2 3x}{cos^2 x} = (\frac{sin 3x}{sin x} - \frac{cos\ 3x}{cos \ x})(\frac{sin 3x}{sin x} + \frac{cos \ 3x}{cos \ x}) \stackrel{(*), (**)}{==} (3-4sin^2x+3-4cos^2x)(3-4sin^2x-3+4cos^2x) =\\\;\\=[6-4(sin^2x+cos^2x)][4(cos^2x-sin^2x)] = (6-4)\cdot4cos2x = 2\cdot4cos2x = 8cos2x .}

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