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Question Number 95001 by mathocean1 last updated on 22/May/20

$$\mathrm{Exercise}$$$$\mathrm{Given}\mathrm{a},\mathrm{b}\in {\mathbb{R}}^{\ast }\mathrm{and}t\mathrm{is}\mathrm{a}\mathrm{variable}\mathrm{real}.$$$$1)\mathrm{Solve}\mathrm{in}{\mathbb{R}}^2\mathrm{for}x,y\mathrm{this}\mathrm{system}:$$$$\{\begin{array}{l}x\sin t-y\cos t=-\mathrm{a}\\ x\cos t+y\sin t=b.\end{array}$$$$$$$$2)/\mathrm{Demonstrate}\mathrm{that}\mathrm{these}\mathrm{solutions}\mathrm{can}$$$$\mathrm{be}\mathrm{written}\mathrm{like}\mathrm{this}(r\mathrm{and}\theta \in \mathbb{R}).$$$$\{\begin{array}{l}x=rcos(t+\theta )\\ y=rsin(t+\theta )\end{array}$$$$$$$$3)\mathrm{we}\mathrm{suppose}\mathrm{now}\mathrm{that}\mathrm{a}=\mathrm{b}=1\mathrm{and}\theta =\frac{\pi }{12}$$$$\mathrm{solve}\mathrm{this}\mathrm{in}[0;2\pi [$$$$\{\begin{array}{l}rcos(t+\theta )⩾-1\\ rsin(t+\theta )<-1\end{array}$$

Answered by bobhans last updated on 22/May/20

$$\left(1\right)\Rightarrow \left(\mathrm{i}\right){\mathrm{x}}^2\sin {}^2\mathrm{t}-2\mathrm{xysin}\mathrm{tcos}\mathrm{t}+{\mathrm{y}}^2\cos {}^2\mathrm{t}={\mathrm{a}}^2$$$$\Rightarrow \left(\mathrm{ii}\right){\mathrm{x}}^2\cos {}^2\mathrm{t}+2\mathrm{xycos}\mathrm{tsin}\mathrm{t}+{\mathrm{y}}^2\sin {}^2\mathrm{t}={\mathrm{b}}^2$$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow x^2+y^2-2xy(\sin t\cos t-\cos t\sin t)=a^2+b^2$$$$\Rightarrow x^2+y^2=a^2+b^2$$

Answered by mathmax by abdo last updated on 22/May/20

$$1)\{\begin{array}{l}\mathrm{sint}\mathrm{x}-\mathrm{cost}\mathrm{y}=-\mathrm{a}\\ \mathrm{cost}\mathrm{x}+\mathrm{sint}\mathrm{y}=\mathrm{b}\end{array}$$$${\mathrm{\Delta }}_{\mathrm{s}}=\left|\begin{array}{c}\mathrm{sint}-\mathrm{cost}\\ \mathrm{cost}\mathrm{sint}\end{array}\right|={\sin }^2\mathrm{t}+{\cos }^2\mathrm{t}?=1\ne 0\Rightarrow$$$$\mathrm{x}=\frac{{\mathrm{\Delta }}_{\mathrm{x}}}{\mathrm{\Delta }}\mathrm{and}\mathrm{y}=\frac{\mathrm{\Delta }\mathrm{y}}{\mathrm{\Delta }}\mathrm{we}\mathrm{have}{\mathrm{\Delta }}_{\mathrm{x}}=\left|\begin{array}{c}-\mathrm{a}-\mathrm{cost}\\ \mathrm{b}\mathrm{sint}\end{array}\right|=-\mathrm{asint}+\mathrm{bcost}$$$${\mathrm{\Delta }}_{\mathrm{y}}=\left|\begin{array}{c}\mathrm{sint}-\mathrm{a}\\ \mathrm{cost}\mathrm{b}\end{array}\right|=\mathrm{bsint}+\mathrm{acost}\Rightarrow$$$$\mathrm{x}=\mathrm{bcost}-\mathrm{asint}\mathrm{and}\mathrm{y}=\mathrm{acost}+\mathrm{bsint}$$$$2)\mathrm{we}\mathrm{have}\mathrm{x}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}(\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{cost}-\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{sint})$$$$\mathrm{let}\cos \theta =\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{and}\sin \theta =\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{and}\mathrm{r}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$$$$\Rightarrow \mathrm{x}=\mathrm{r}(\mathrm{cost}\cos \theta -\mathrm{sint}\sin \theta )=\mathrm{r}\cos (\mathrm{t}+\theta )$$$$\mathrm{y}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{cost}+\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{sint})=\mathrm{r}(\sin \theta \mathrm{cost}+\cos \theta \mathrm{sint})$$$$=\mathrm{r}\sin (\mathrm{t}+\theta )$$$$$$

Commented by mathocean1 last updated on 08/Jun/20

$$thankssir$$

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