изпит от ту

[Ивана К.]

71333150_548346099317193_4955099401413459968_n.png (16.99 KiB) Прегледано 536 пъти

[Добромир Глухаров]

$A=sin^2(x+y)cos(x-y)+sin^2(x-y)cos(x+y)$

Знаем, че $sin(\alpha\pm\beta)=sin\alpha cos\beta\pm cos\alpha sin\beta\Rightarrow sin(\alpha+\beta)+sin(\alpha-\beta)=2sin\alpha cos\beta$

$\Rightarrow sin\alpha cos\beta=\frac{1}{2}[sin(\alpha+\beta)+sin(\alpha-\beta)]$

$\Rightarrow sin(x+y)cos(x-y)=\frac{1}{2}(sin2x+sin2y),sin(x-y)cos(x+y)=\frac{1}{2}(sin2x-sin2y)$

$A=sin(x+y)\cdot\frac{1}{2}(sin2x+sin2y)+sin(x-y)\cdot\frac{1}{2}(sin2x-sin2y)$

$2A=sin2x(sin(x+y)+sin(x-y))+sin2y(sin(x+y)-sin(x-y))=sin2x.2sin\frac{(x+y)+(x-y)}{2}cos\frac{(x+y)-(x-y)}{2}+sin2y.2sin\frac{(x+y)-(x-y)}{2}cos\frac{(x+y)+(x-y)}{2}$

$A=sin2xsinxcosy+sin2ysinycosx=2sin^2xcosxcosy+2sin^2ycosycosx=2cosxcosy(sin^2x+sin^2y)=3cosxcosy$

$sin^2x+sin^2y=\frac{3}{2}\Rightarrow2-cos^2x-cos^2y=\frac{3}{2}$

$cos^2x+cos^2y=\frac{1}{2};A=3cosxcosy$

Полагаме $cos^2x=t\in[0;1]\Rightarrow cos^2y=\frac{1}{2}-t$

$A=3cosxcosy=\pm3\sqrt{t}\sqrt{\frac{1}{2}-t}=\pm3\sqrt{\varphi(t)}$

$\varphi(t)=t\left(\frac{1}{2}-t\right)>0\Rightarrow t\in\left[0;\frac{1}{2}\right]$

$\min_{t\in\left[0;\frac{1}{2}\right]}\varphi(t)=\varphi(0)=\varphi\left(\frac{1}{2}\right)=0$

$\max_{t\in\left[0;\frac{1}{2}\right]}\varphi(t)=\varphi\left(\frac{1}{4}\right)=\frac{1}{16}$

$minA=-3\sqrt{\frac{1}{16}}=-\frac{3}{4}$

$maxA=3\sqrt{\frac{1}{16}}=\frac{3}{4}$