Explicație pas cu pas:
[tex]\sin^{2}x+\cos^{2}x= 1[/tex]
a) folosim formula:
[tex] {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )[/tex]
deci:
[tex]\sin^{3}x + \cos^{3}x = (\sin\: x + \cos\: x)(\sin^{2}x - \sin\: x\cos\: x + \cos^{2}x = (\sin\: x+ \cos\: x)(1 - \sin\: x\cos\: x)[/tex]
b) folosim formula:
[tex]{a}^{4} + {b}^{4} = ({a}^{2} + {b}^{2})^{2} - 2a^{2} b^{2}[/tex]
deci:
[tex] \sin^{4} (x) + \cos^{4}(x) = (\sin^{2}(x) + \cos^{2}(x))^{2} - 2\sin^{2}(x)\cos^{2}(x) = (1)^{2} - 2\sin^{2}(x)\cos^{2}(x) = 1 - 2\sin^{2}(x)\cos^{2}(x)[/tex]
c) folosim formula:
[tex]{a}^{3} + {b}^{3} = (a + b)^{3} - 3ab(a + b)[/tex]
deci:
[tex]( { \sin}^{2}x )^{3} + ( { \cos}^{2}x)^{3} = ({ \sin}^{2}x+ { \cos}^{2}x)^{3} - 3\sin^{2}x \times \cos^{2}x\times ( \sin^{2}x + \cos^{2}x) = (1)^{3} - 3\sin^{2}x \times \cos^{2}x\times (1) = 1 - 3\sin^{2}x\cos^{2}x[/tex]
