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Baiatul122001
a fost răspuns

Aratati ca:
a)[tex] \frac{1- tg^{2}x }{1+ tg^{2}x }=cos2x [/tex]
b)[tex] \frac{sin4x}{cos2x+ cos^{2}2x }=2tgx [/tex]

Răspuns :

GreenEyes71
Salut,

[tex]\dfrac{1-tg^2x}{1+tg^2x}=\dfrac{1-tg^2x}{1+\dfrac{sin^2x}{cos^2x}}=\dfrac{1-tg^2x}{\dfrac{cos^2x+sin^2x}{cos^2x}}=cos^2x(1-tg^2x)=\\\\\\=cos^2x-cos^2x\cdot tg^2x=cos^2x-sin^2x=cos(2x).[/tex]

Simplu, nu ? :-).

Green eyes.

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