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salut . sin3x+sinx supra cos3x+cosx

Răspuns :

Salut,


[tex]\dfrac{sin3x+sinx}{cos3x+cosx}=\dfrac{2\cdot sin\dfrac{3x+x}2\cdot cos\dfrac{3x-x}2}{2\cdot cos\dfrac{3x+x}2\cdot cos\dfrac{3x-x}2}=tg2x.[/tex]


Green eyes.