Rezolvare:
Utilizăm formulele de calcul prescurtat:
[tex](a + b)^2 = a^2 +2ab + b^2 \Rightarrow a^2 + b^2 = (a + b)^2 - 2ab\\[/tex]
[tex]\sin^{4} t + \cos^{4} t - 1 = (\sin^{2} t)^2 + (\cos^{2} t)^2 - 1 = \\[/tex]
Utilizăm identitatea:
[tex]\boxed{\boldsymbol{ \sin^{2} \alpha + \cos^{2} \alpha = 1 }}[/tex]
[tex]= \big(\underbrace{\sin^{2} t + \cos^{2} t}_{1}\big)^2 - 2\sin^{2} t \cos^{2} t \\[/tex]
[tex]= 1 - 2\sin^{2} t \cos^{2} t + 1 = - 2\sin^{2} t \cos^{2} t[/tex]
Utilizăm formulele de calcul prescurtat:
[tex]a^3 + b^3 = (a + b)^{3} - 3ab(a + b)[/tex]
[tex]\sin^{6} t + \cos^{6} t - 1 = (\sin^{2} t)^3 + (\cos^{2} t)^3 - 1 =\\[/tex]
[tex]= \big(\underbrace{\sin^{2} t + \cos^{2} t}_{1}\big)^3 - 3\sin^{2} t \cos^{2} t \big(\underbrace{\sin^{2} t + \cos^{2} t}_{1}\big) \\[/tex]
[tex]= 1 - 3\sin^{2} t \cos^{2} t + 1 = - 3\sin^{2} t \cos^{2} t\\[/tex]
Egalitatea devine:
[tex]\dfrac{\sin^{4} t + \cos^{4} t - 1}{\sin^{6} t + \cos^{6} t - 1} = \dfrac{- 2\sin^{2} t \cos^{2} t}{- 3\sin^{2} t \cos^{2} t} =\bf \dfrac{2}{3}\\[/tex]
q.e.d.
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