14)
Formula lui De Moivre:
[tex](\cos x+i\sin x)^n = \cos (nx) + i\sin (nx)[/tex]
[tex](\cos \alpha -i\sin \alpha )(\cos 5\alpha +i\sin 5\alpha ) = \\ \\ = (\cos \alpha -i\sin \alpha )(\cos \alpha +i\sin \alpha ) ^5 = \\ \\ =(\cos \alpha -i\sin \alpha )(\cos \alpha +i\sin \alpha )(\cos \alpha +i\sin \alpha )^4 = \\ \\ = (\cos^2\alpha +\sin^2 \alpha )(\cos \alpha +i\sin \alpha )^4 = \\ \\ = 1\cdot (\cos \alpha +i\sin \alpha )^4 = (\cos \alpha +i\sin \alpha )^4= \\ \\ =\cos4\alpha +i \sin 4\alpha \in \mathbb{R}[/tex]
[tex]\Rightarrow \sin 4\alpha = 0 \Rightarrow \alpha = \dfrac{k\pi}{4},\quad k\in \mathbb{Z}[/tex]
⇒ B) corect
15)
[tex]A(4,0);\quad B(0,3)\\ \\ d:\quad x+y = 0 \\ \\ M(a,b)\in d\Rightarrow a+b = 0\Rightarrow a = -b \\ \\ MA^2+MB^2 = (a-4)^2+b^2+a^2+(b-3)^2\\ \\ = (-b-4)^2+b^2+(-b)^2+(b-3)^2 = \\ \\ =(b+4)^2+(b+0)^2+(b+0)^2+(b-3)^2[/tex]
Observăm că această expresie este o parabolă.
Centralizăm expresia, iar minimul său va fi chiar în centrul expresiei.
[tex] b = \dfrac{-4+0+0+3}{4} = -\dfrac{1}{4}\\ \\\\ \Rightarrow \Big(-\dfrac{1}{4}+4\Big)^2+\dfrac{1}{16}+\dfrac{1}{16}+\Big(-\dfrac{1}{4}-3\Big)^2 = \\ \\\\ = \dfrac{225+2+169}{16}=\dfrac{99}{4}[/tex]
=> A) corect[tex][/tex]
