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Question Number 38151 by ajfour last updated on 22/Jun/18

$$a>b>0$$ $$a^2\cos \theta -b^2\sin \theta =(a^2-b^2)\sin \theta \cos \theta$$ $$Find\theta intermsofa,b.$$ $$\theta \in (0,\frac{\pi }{2}).$$

Commented bybehi83417@gmail.com last updated on 23/Jun/18

$$let:m=e^{i\theta },i=\sqrt{-1}$$ $$\Rightarrow a^2.\frac{m^2+1}{2m}-b^2.\frac{m^2-1}{2im}=(a^2-b^2)\frac{m^4-1}{4{im}^2}$$ $$\Rightarrow 2a^2.m(m^2+1)+2{mib}^2(m^2-1)=i(b^2-a^2)(m^4-1)$$ $$\Rightarrow \{\begin{array}{l}2a^2.m(m^2+1)=0\left(A\right)\\ 2b^2.m(m^2-1)=(b^2-a^2)(m^4-1)\left(B\right)\end{array}$$ $$\stackrel{A}{\Rightarrow }(a=0)\vee (m=0)\vee (m^2+1)=0$$ $$a=0\Rightarrow cos\theta =1\Rightarrow \theta =0\left(\times \right)$$ $$m=0\Rightarrow e^{i\theta }=0\left(\times \right)$$ $$m^2+1=0\Rightarrow m=\pm i\Rightarrow e^{i\theta }=e^{\pm i\frac{\pi }{2}}\Rightarrow \theta =\pm \frac{\pi }{2}\left(\times \right)$$ $$\stackrel{B}{\Rightarrow }2b^2.m(m^2-1)=(b^2-a^2)(m^4-1)$$ $$\Rightarrow \{\begin{array}{l}{m}^2-1=0\Rightarrow m=\pm 1\Rightarrow e^{i\theta }=\pm 1=e^{\pm i\pi }\\ \Rightarrow \theta =\pm \pi \left(\times \right)\end{array}\left(\times \right)$$ $$2b^2.m=(b^2-a^2)(m^2+1)\Rightarrow \frac{m^2+1}{2m}=\frac{b^2}{b^2-a^2}$$ $$\frac{m^2+1+2m}{m^2+1-2m}=\frac{b^2+b^2-a^2}{b^2-b^2+a^2}\Rightarrow {\left(\frac{m+1}{m-1}\right)}^2=2\frac{b^2}{a^2}-1$$ $$\Rightarrow \frac{m+1}{m-1}=\frac{\sqrt{2b^2-a^2}}{a}\Rightarrow \frac{m+1+m-1}{m+1-m+1}=\frac{\sqrt{2b^2-a^2}+a}{\sqrt{2b^2-a^2}-a}$$ $$\Rightarrow \mathit{m}=\frac{{\mathit{b}}^2+\mathit{a}.\sqrt{2{\mathit{b}}^2-{\mathit{a}}^2}}{{\mathit{b}}^2-{\mathit{a}}^2}$$ $$\Rightarrow {\mathit{e}}^{\mathit{i}\mathit{\theta }}=\frac{{\mathit{b}}^2+\mathit{a}.\sqrt{2{\mathit{b}}^2-{\mathit{a}}^2}}{{\mathit{b}}^2-{\mathit{a}}^2},(\mathit{p}=\frac{{\mathit{b}}^2}{{\mathit{a}}^2})$$ $$\Rightarrow \mathit{i}\mathit{\theta }=\mathit{l}\mathit{n}\frac{{\mathit{b}}^2+\mathit{a}\sqrt{2{\mathit{b}}^2-{\mathit{a}}^2}}{{\mathit{b}}^2-{\mathit{a}}^2}\Rightarrow \mathit{i}\mathit{\theta }=\mathit{l}\mathit{n}\frac{\mathit{p}+\sqrt{2\mathit{p}-1}}{\mathit{p}-1}.◼$$ $$=\mathit{l}\mathit{n}(\frac{\mathit{p}}{\mathit{p}-1}+\frac{\sqrt{1-2\mathit{p}}}{\mathit{p}-1}.\mathit{i})=$$ $$=[\frac{1}{2}\mathit{l}\mathit{n}\left(\frac{{\mathit{p}}^2+(1-2\mathit{p})}{{(\mathit{p}-1)}^2}\right)+\mathit{i}.\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\frac{\frac{\sqrt{1-2\mathit{p}}}{\mathit{p}-1}}{\frac{\mathit{p}}{\mathit{p}-1}}]=$$ $$=[\frac{1}{2}\mathit{l}\mathit{n}1+\mathit{i}.\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\frac{\sqrt{1-2\mathit{p}}}{\mathit{p}}]=\mathit{i}.\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\frac{\sqrt{1-2\mathit{p}}}{\mathit{p}}$$ $$\Rightarrow \mathit{\theta }=\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\frac{\sqrt{1-2\mathit{p}}}{\mathit{p}}=\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{g}\frac{\mathit{a}.\sqrt{{\mathit{a}}^2-2{\mathit{b}}^2}}{{\mathit{b}}^2}.◼$$ $$\mathit{s}\mathit{i}\mathit{r}\mathit{A}\mathit{j}\mathit{f}\mathit{o}\mathit{u}\mathit{r}!\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{a}\mathit{r}\mathit{e}\mathit{a}\mathit{l}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$ $$\mathit{f}\mathit{o}\mathit{r}\mathit{y}\mathit{o}\mathit{u}\mathit{r}\mathit{q}\mathit{u}\mathit{i}\mathit{s}\mathit{t}\mathit{i}\mathit{o}\mathit{n}.$$

Commented byajfour last updated on 23/Jun/18

$$therehastobearealsolution..$$

Commented byMJS last updated on 23/Jun/18

$$\mathrm{the}\mathrm{mistake}\mathrm{is}\mathrm{simple}\mathrm{but}\mathrm{it}\mathrm{took}\mathrm{me}\mathrm{quite}$$ $$\mathrm{some}\mathrm{time}\mathrm{to}\mathrm{see}\mathrm{it}:$$ $$2a^{2}m(m^2-1)=0\wedge 2b^{2}m(m^2-1)=(b^2-a^2)(m^4-1)$$ $$\mathrm{you}\mathrm{solved}\mathrm{the}\mathrm{second}\mathrm{term}\mathrm{but}\mathrm{the}\mathrm{first}\mathrm{term}$$ $$\mathrm{is}\ne 0,\mathrm{so}\mathrm{the}\mathrm{whole}\mathrm{equation}\mathrm{is}not\mathrm{solved}$$

Commented bybehi83417@gmail.com last updated on 23/Jun/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$ $$2a^2.m(m^2+1)=0,andithassolved.$$

Commented byMJS last updated on 23/Jun/18

$$\mathrm{sorry},\mathrm{typo}.$$ $$\mathrm{but}\mathrm{have}\mathrm{you}\mathrm{tried}\mathrm{your}\mathrm{solution}\mathrm{with}\mathrm{any}$$ $$\mathrm{real}\mathrm{numbers}a\mathrm{and}b?$$ $$a=5$$ $$b=3$$ $$\arctan \frac{3^2}{\sqrt{5^2-2\times 3^2}}=\arctan \frac{9\sqrt{7}}{7}$$ $$25\cos \arctan \frac{9\sqrt{7}}{7}-9\sin \arctan \frac{9\sqrt{7}}{7}=$$ $$=\frac{25\sqrt{154}}{44}-\frac{81\sqrt{22}}{44}\approx -1.58368$$ $$16\sin \arctan \frac{9\sqrt{7}}{7}\times \cos \arctan \frac{9\sqrt{7}}{7}=\frac{18\sqrt{7}}{11}\approx 4.32941$$ $$\mathrm{so}\mathrm{something}\mathrm{went}\mathrm{wrong}$$ $$$$ $$25\cos \theta -9\sin \theta =16\sin \theta \cos \theta \mathrm{has}\mathrm{the}$$ $$\mathrm{approximate}\mathrm{solution}0.93305,\mathrm{while}$$ $$\arctan \frac{9\sqrt{7}}{7}\approx 1.2849$$

Commented byMJS last updated on 23/Jun/18

$$\mathrm{sorry}\mathrm{I}\mathrm{misread}$$ $$\mathrm{still}\mathrm{with}\mathrm{this}\mathrm{value}\mathrm{the}\mathrm{left}\mathrm{side}\mathrm{gives}$$ $$-26.523$$ $$\mathrm{but}\mathrm{the}\mathrm{righr}\mathrm{side}\mathrm{is}$$ $$-4.3294$$ $$\mathrm{and}\mathrm{the}\mathrm{new}\theta \approx 2.8557\mathrm{while}\mathrm{the}\mathrm{solution}$$ $$\mathrm{is}\approx .93305$$

Commented bybehi83417@gmail.com last updated on 23/Jun/18

$$dearMJS!pleasecorrectmytypoor$$ $$postcorrectanswer.$$ $$iknowyouranswerwillthebest,finally.$$

Commented byMJS last updated on 24/Jun/18

$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{mean}\mathrm{to}\mathrm{offend}\mathrm{you},\mathrm{I}\mathrm{just}\mathrm{wanted}\mathrm{to}$$ $$\mathrm{say}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{some}\mathrm{mistake}.\mathrm{the}\mathrm{solution}\mathrm{must}$$ $$\mathrm{solve}\mathrm{the}\mathrm{equation}...\mathrm{we}\mathrm{all}\mathrm{should}\mathrm{test}\mathrm{our}$$ $$\mathrm{solutions}\mathrm{with}\mathrm{at}\mathrm{least}2\mathrm{or}3\mathrm{sets}\mathrm{of}\mathrm{real}$$ $$\mathrm{numbers}\mathrm{before}\mathrm{posting}\mathrm{them}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

$$t=tan\frac{\theta }{2}$$ $$a^2\left(\frac{1-t^2}{1+t^2}\right)-b^2\left(\frac{2t}{1+t^2}\right)=(a^2-b^2)\frac{2t}{1+t^2}\times \frac{1-t^2}{1+t^2}$$ $$\frac{a^2-a^2{t}^2-2b^{2}t}{1+t^2}=\frac{(a^2-b^2)(2t-2t^3)}{(1+t^2)(1+t^2)}$$ $$(1+t^2)(a^2-a^2{t}^2-2b^{2}t)=(2a^{2}t-2a^2{t}^3-2b^{2}t+$$ $$2b^2{t}^3)$$ $$a^2-a^2{t}^2-2b^{2}t+a^2{t}^2-a^2{t}^4-2b^2{t}^3=$$ $$2a^{2}t-2a^2{t}^3-2b^{2}t+2b^2{t}^3$$ $$a^2-a^2{t}^4-2b^2{t}^3-2a^{2}t+2a^2{t}^3+2b^{2}t-2b^2{t}^3=0$$ $$t^4(-a^2)+t^3(2a^2-4b^2)+t(-2a^2+2b^2)+a^2=0$$ $$t_1+t_2+t_3+t_4=\frac{-(2a^2-4b^2)}{-a^2}$$ $$\mathrm{\Sigma }{t}_1{t}_2=0$$ $$\mathrm{\Sigma }{t}_1{t}_2{t}_3=-\frac{(-2a^2+2b^2)}{-a^2}$$ $$t_1{t}_2{t}_3{t}_4=\frac{a^2}{-a^2}$$ $$wait...$$ $$$$

Commented bymath khazana by abdo last updated on 22/Jun/18

$$butthetraindontwait....$$

Commented byrahul 19 last updated on 22/Jun/18

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Answered by ajfour last updated on 23/Jun/18

$$letfortheellipseofQ.38092$$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ $$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}\left(\frac{b^2}{a^2}\right)$$ $$slopeofnormaltocircleat$$ $$pointoftangency$$ $$\left(\frac{r-y}{r-x}\right)=\frac{y}{x}\left(\frac{a^2}{b^2}\right)$$ $$letr-x=r\cos \theta$$ $$r-y=r\sin \theta$$ $$\Rightarrow \frac{\sin \theta }{\cos \theta }=\left(\frac{1-\sin \theta }{1-\cos \theta }\right)\left(\frac{a^2}{b^2}\right)$$ $$a^2\cos \theta (1-\sin \theta )=b^2\sin \theta (1-\cos \theta )$$ $$a^2\sin \theta -b^2\sin \theta =(a^2-b^2)\sin \theta \cos \theta$$ $$...$$ $$$$ $$$$

Answered by MJS last updated on 25/Jun/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{nice}\mathrm{and}\mathrm{probably}\mathrm{not}\mathrm{useable}\mathrm{but}$$ $${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}$$ $$\mathrm{let}a=1\mathrm{and}{b}^2=q(q=\frac{b^2}{a^2})$$ $$\cos \theta -q\sin \theta -(1-q)\sin \theta \cos \theta =0$$ $$\theta =2\arctan t$$ $$-\frac{t^4+2(2q-1)t^3+2t-1}{t^4+2t^2+1}=0$$ $$t^4+2t+1=0\Rightarrow t=\pm \mathrm{i}\notin \mathbb{R}\Rightarrow \mathrm{we}\mathrm{can}\mathrm{multiplicate}$$ $$\mathrm{without}\mathrm{further}\mathrm{trouble}$$ $$t^4+2(2q-1)t^3+2t-1=0$$ $$\mathrm{approximate}\mathrm{solving}\mathrm{with}\mathrm{some}\mathrm{values}\mathrm{of}q$$ $$\mathrm{leads}\mathrm{to}\mathrm{the}\mathrm{conclusion}\mathrm{that}\mathrm{this}\mathrm{has}\mathrm{two}$$ $$\mathrm{complex}\mathrm{and}\mathrm{two}\mathrm{real}\mathrm{solutions}.\mathrm{the}\mathrm{complex}$$ $$\mathrm{ones}\mathrm{must}\mathrm{be}\alpha +\beta \mathrm{i}\mathrm{and}\alpha -\beta \mathrm{i}\mathrm{and}\mathrm{I}\mathrm{hope}\mathrm{the}$$ $$\mathrm{real}\mathrm{ones}\mathrm{are}\gamma +\delta \mathrm{and}\gamma -\delta .$$ $$(t-\alpha -\beta \mathrm{i})(t-\alpha +\beta \mathrm{i})(t-\gamma -\delta )(t-\gamma +\delta )=0$$ $$t^4-$$ $$-2(\alpha +\gamma )t^3+$$ $$+({\alpha }^2+4\vartheta \gamma +{\beta }^2+{\gamma }^2-{\delta }^2)t^2-$$ $$-2({\alpha }^2\gamma +\alpha {\gamma }^2-\alpha {\delta }^2+{\beta }^2\gamma )t+$$ $$+({\alpha }^2+{\beta }^2)({\gamma }^2-{\delta }^2)=0$$ $$\Rightarrow$$ $$\left[1\right]2(2q-1)=-2(\alpha +\gamma )$$ $$\left[2\right]0={\alpha }^2+4\vartheta \gamma +{\beta }^2+{\gamma }^2-{\delta }^2$$ $$\left[3\right]2=-2({\alpha }^2\gamma +\alpha {\gamma }^2-\alpha {\delta }^2+{\beta }^2\gamma )$$ $$\left[4\right]-1=({\alpha }^2+{\beta }^2)({\gamma }^2-{\delta }^2)$$ $$$$ $$\left[1\right]\alpha =-\gamma -2q+1$$ $$\left[2\right]\beta =\sqrt{-{\alpha }^2-4\alpha \gamma -{\gamma }^2+{\delta }^2}=$$ $$=\sqrt{2{\gamma }^2+(4q-2)\gamma +{\delta }^2-{(2q-1)}^2}$$ $$\left[3\right]2=-4{\gamma }^3+(6-12q){\gamma }^2-2(2\gamma +2q-1){\delta }^2\Rightarrow$$ $$\Rightarrow \delta =\sqrt{-\frac{2{\gamma }^3+3(2q-1){\gamma }^2+1}{2\gamma +2q-1}}$$ $$\Rightarrow \beta =\sqrt{\frac{2{\gamma }^3+3(2q-1){\gamma }^2-2q(4q^2-6q+3)}{2\gamma +2q-1}}$$ $$\left[4\right]-1=\frac{16{\gamma }^6+48(2q-1){\gamma }^5+48{(2q-1)}^2{\gamma }^4+16{(2q-1)}^3{\gamma }^3+4(2q-1){\gamma }^2+4{(2q-1)}^2\gamma -1}{{(2\gamma +2q-1)}^2}$$ $$[\gamma \ne \frac{1}{2}-q;\mathrm{this}\mathrm{will}\mathrm{be}\mathrm{fulfilled}\mathrm{with}$$ $$q=\frac{1-\sqrt[3]{2}}{2}<0\mathrm{so}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{to}\mathrm{care}]$$ $${\gamma }^6+3(2q-1){\gamma }^5+3{(2q-1)}^2{\gamma }^4+{(2q-1)}^3{\gamma }^3+\frac{q}{2}{\gamma }^2+\frac{q(2q-1)}{2}\gamma +\frac{q(q-1)}{4}=0$$ $$\mathrm{at}\mathrm{this}\mathrm{point}\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{try}\mathrm{to}\mathrm{omit}\mathrm{the}{\gamma }^5$$ $$\mathrm{by}\mathrm{setting}\gamma =u-\frac{3(2q-1)}{6}=u-q+\frac{1}{2}$$ $$\mathrm{which}\mathrm{leads}\mathrm{to}$$ $$u^6-\frac{3{(2q-1)}^2}{4}{u}^4+\frac{48q^4-96q^3+72q^2-16q+3}{16}{u}^2-\frac{{(8q^3-12q^2+6q+1)}^2}{64}=0$$ $$\mathrm{we}\mathrm{seem}\mathrm{to}\mathrm{have}\mathrm{good}\mathrm{luck}$$ $$\mathrm{set}u=\sqrt{v}$$ $$v^3-\frac{3{(2q-1)}^2}{4}{v}^2+\frac{48q^4-96q^3+72q^2-16q+3}{16}v-\frac{{(8q^3-12q^2+6q+1)}^2}{64}=0$$ $$\mathrm{omit}{v}^2\mathrm{by}\mathrm{setting}v=w-\frac{{(2q-1)}^2}{4}\mathrm{leads}\mathrm{to}$$ $$w^3+\frac{q}{2}w+\frac{q(q-1)}{4}=0$$ $$[x^3+\mathcal{P}x+\mathcal{Q}=0\Rightarrow$$ $$\Rightarrow x_1=\sqrt[3]{-\frac{\mathcal{Q}}{2}+\sqrt{\frac{{\mathcal{P}}^3}{27}+\frac{{\mathcal{Q}}^2}{4}}}+\sqrt[3]{-\frac{\mathcal{Q}}{2}-\sqrt{\frac{{\mathcal{P}}^3}{27}+\frac{{\mathcal{Q}}^2}{4}}}]$$ $$w_1=\frac{\sqrt[3]{q}}{2}(\sqrt[3]{1-q+\sqrt{{(q-1)}^2+\frac{8q}{27}}}+\sqrt[3]{1-q-\sqrt{{(q-1)}^2+\frac{8q}{27}}})$$ $${(q-1)}^2+\frac{8q}{27}>0\mathrm{\forall }q>0\Rightarrow w_1\in \mathbb{R}\mathrm{\forall }q>0$$ $$...\mathrm{now}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{set}q\mathrm{and}\mathrm{go}\mathrm{all}\mathrm{the}\mathrm{way}$$ $$\mathrm{back}...\mathrm{solutions}\mathrm{of}\mathrm{polynomes}\mathrm{of}{4}^{\mathrm{th}}\mathrm{degree}$$ $$\mathrm{usually}\mathrm{are}\mathrm{even}\mathrm{worse}...$$

Commented byajfour last updated on 25/Jun/18

$$\mathcal{A}wesome!manymanythanksSir.$$

Commented byMJS last updated on 25/Jun/18

$$\mathrm{you}{\mathrm{won}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{able}\mathrm{to}\mathrm{use}\mathrm{this}\mathrm{for}\mathrm{further}$$ $$\mathrm{simplifications}\mathrm{or}\mathrm{transformations}\mathrm{but}\mathrm{at}$$ $$\mathrm{least}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{calculate}\mathrm{approximate}$$ $$\mathrm{values}\mathrm{using}\mathrm{a}\mathrm{calculator}...$$ $$\mathrm{by}\mathrm{the}\mathrm{way},\mathrm{it}\mathrm{seems}\beta \mathrm{and}\delta \mathrm{are}\mathrm{imaginary}$$ $$\mathrm{numbers},\mathrm{so}\alpha \pm \beta \mathrm{i}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\gamma \pm \delta \mathrm{will}$$ $$\mathrm{be}\mathrm{complex}...$$

Commented byMJS last updated on 25/Jun/18

$$...\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{try}{w}_2\mathrm{and}{w}_3,\mathrm{if}\mathrm{you}\mathrm{like}\mathrm{to}:$$ $$w_1=\sqrt[3]{\mathcal{A}}+\sqrt[3]{\mathcal{B}}$$ $$\Rightarrow w_2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})\sqrt[3]{\mathcal{A}}+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})\sqrt[3]{\mathcal{B}}$$ $$w_3=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})\sqrt[3]{\mathcal{A}}+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})\sqrt[3]{\mathcal{B}}$$

Commented byajfour last updated on 26/Jun/18

$$Igotitreducedtoacubiceq.$$ $$quiteeasily,ishallsharethat$$ $$soon..$$

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