Question

Matematica va rog ajutatima

Answer 1

 

$\displaystyle\bf\\1)\\\frac{\pi}{9}=\frac{180}{9}=20^o \in Cadranul~1\\\\-\frac{\pi}{9}=-\frac{180}{9}=-20^o \in Cadranul~4\\\\Cosinusul~este~pozitiv~in~cadranele~1~si~4\\\\Avem~formula~~cos(-x)=cos(x)\\\\\implies~cos(-20^o)=cos(20^o)\\\\\implies~\boxed{\bf cos\Big(-\frac{\pi}{4}\Big)=cos\Big(\frac{\pi}{4}\Big)}$

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$\displaystyle\bf\\2)\\f:D\to R,~f(x)=7^{\sqrt{1-2x}}\\\\Conditie: Sub~radical~nu~este~admis~numar~negativ:\\\\1-2x\geq0\\\\-2x\geq -1~~\Big|\cdot(-1)\\\\\textbf{Schimbam sensul inegalitatii.}\\\\2x\leq1\\\\x\leq\frac{1}{2}\\\\ \boxed{\bf D=\Big(-\infty,~\frac{1}{2}\Big]}$

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$\displaystyle\bf\\3)\\\textbf{Daca o dreapta este tangenta la un cerc, atunci raza cercului}\\\textbf{dusa din cetru in punctul de tangenta este perpendiculara pe}\\\textbf{tangenta la cerc.}\\\\\implies~m(\sphericalangle OBC)=90^o\\\Delta OAB~este~triunghi~isoscel~deoarece~OA=OB~fiind~razele~cercului.\\\\\implies~m(\sphericalangle OAB)=m(\sphericalangle OBA)=\frac{180^o-100^o}{2}= \frac{80^o}{2}=40^o\\\\m(\sphericalangle ABC)=m(\sphericalangle OBC)-m(\sphericalangle OBA)=90^o-40^o=\boxed{\bf50^o}$

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$\displaystyle\bf\\4)\\E=(sin~x-cos~x)^2+sin~2x\\\\Folosim~formulele:\\\\(a-b)^2=a^2-2ab+b^2\\\\sin^2x+cos^2x=1\\\\sin~2x=2sin~x~cos~x\\\\Rezolvare:\\\\E=sin^2x-2sin~x~cos~x+cos^2x+sin~2x\\\\E=sin^2x+cos^2x-2sin~x~cos~x+2sin~x~cos~x\\\\E=1-2sin~x~cos~x+2sin~x~cos~x\\\\\boxed{\bf E=1}$