Question

ma ajutati si pe mine cu tema sunt niste calcule de care nu ma prind

Answer 1

Formule de calcul prescurtat:

$a^2+b^2 = (a+b)^2 - 2ab\\ a^3+b^3 = (a+b)(a^2-ab+b^2)$

Rezolvare:

$\sin x\cdot \cos x = \dfrac{1}{3}\\\\\sin^4 x+\cos^4 x = (\sin^2 x)^2+(\cos^2 x)^2 =\\ \\ =(\sin^2x+\cos^2 x)^2 - 2\sin^2 x\cos^2 x =\\ \\ =1^2 - 2(\sin x\cos x)^2 = \\ \\ =1 - 2\cdot \Big(\dfrac{1}{3}\Big)^2 = \boxed{\dfrac{7}{9}}\\ \\ \\ \sin^6 x+\cos^6x =(\sin^2 x)^3+(\cos^2 x)^3 =\\ \\ = (\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x) =\\ \\ = 1\cdot \Big[\sin^4 x+\cos^4 x-(\sin x\cos x)^2\Big] =\\ \\ = \dfrac{7}{9}-\dfrac{1}{9} = \dfrac{6}{9} = \\ \\ =\boxed{\dfrac{2}{3}}$

Answer 2

$(\sin^2x +\cos^2 x)^2=\sin ^4 x+2\cdot\sin^2x\cdot\cos^2 x+\cos ^4 x=\\ \\ \Rightarrow 1^2=\sin ^4 x+\cos ^4 x+ 2\cdot (\sin x \cdot \cos x)^2\\ \\ \Rightarrow 1=\sin ^4 x+\cos ^4 x +2\cdot (\frac{1}{3})^2 \\ \\ \Rightarrow \sin ^4 x+\cos ^4 x=1-\frac{2}{9}\\ \\ \Rightarrow \sin ^4 x+\cos ^4 x=\frac{9}{9}-\frac{2}{9}\\ \\ \Rightarrow \sin ^4 x+\cos ^4 x=\frac{7}{9}$

$\sin ^6 x+\cos ^6 x=(\sin ^2 x)^3+(\cos ^2 x)^3=\\ \\ =(\sin^2 x+\cos^2 x)(\sin ^4 x - \sin^2 x \cos^2 x+\cos^4 x)\\ \\ =1[ \frac{7}{9}- (\sin x\cdot \cos x )^2]= \\ \\ =\frac{7}{9}-(\frac{1}{3})^2=\frac{7}{9}-\frac{1}{9}=\frac{6}{9}=\frac{2}{3}$