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Question Number 35844 by behi83417@gmail.com last updated on 24/May/18

Commented by abdo mathsup 649 cc last updated on 30/May/18

$$letput\theta =arctantwitht>0$$$$\left(e\right)\iff 1+sin\left(arctant\right)=4t\iff$$$$1+\frac{t}{\sqrt{1+t^2}}=4t\iff t+\sqrt{1+t^2}=4t\sqrt{1+t^2}$$$$\iff t=(4t-1)\sqrt{1+t^2}$$$$\iff t^2={(4t-1)}^2(1+t^2)$$$$\iff t^2=(16t^2-8t+1)(1+t^2)$$$$\iff t^2=16t^2-8t+1+16t^4-8t^3+t^2$$$$\iff 16t^4-8t^3+16t^2-8t+1=0...becontinued...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/May/18

$$t=tan\frac{\theta }{2}$$$$1+\frac{2t}{1+t^2}=4\times \frac{2t}{1-t^2}$$$$onsimplification$$$$t^4+10t^3+6t-1=0$$$$(t^2+At+1)(t^2+Bt-1)=0$$$$t^4+{Bt}^3-t^2+{At}^3+{ABt}^2-At+t^2+Bt-1=0$$$$t^4+t^3(A+B)+t^2(-1+AB+1)+t(-A+B)-1=0$$$$t^4+t^3(A+B)+t^2\left(AB\right)+t(-A+B)-1=0$$$$soA+B=10$$$$B-A=6$$$$ABcannotbezero$$$$B=8A=2$$$$(t^2+2t+1)(t^2+8t-1)=0$$$$(t^2+2t+1)cannotbezero$$$$t^2+8t-1=0$$$$t=\frac{-8\pm \sqrt{68}}{2}$$$$=-4\pm \sqrt{17}$$$$\theta liesinfirstquadrantso$$$$t=-4+\sqrt{17}$$$$tan\frac{\theta }{2}=-4+\sqrt{17}$$$$\theta =2\times {tan}^{-1}(-4+\sqrt{17})$$

Commented by Rasheed.Sindhi last updated on 25/May/18

$$\vee \in \mathbb{R}\lambda \mathbb{N}i\mathbb{C}e!$$

Commented by behi83417@gmail.com last updated on 25/May/18

$$(t^2+2t+1)(t^2+8t-1)=t^4+8t^3-t^2+2t^3+$$$$+16t^2-2t+t^2+8t-1=t^4+10t^3+16t^2+6t-1$$

Commented by ajfour last updated on 25/May/18

$$ExcellentSir,ihadthisinmind,$$$${couldn}^{\prime }tfindthetimencourage.$$

Commented by behi83417@gmail.com last updated on 25/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 25/May/18

$$hanks...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/May/18

$$thismethodofsolvemaybetheparadox$$$$ofmathematicsasMr.Behidetectedtheflaw$$

Commented by Rasheed.Sindhi last updated on 26/May/18

$$\mathrm{tanmay}\mathrm{Sir}$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{that}\mathrm{your}\mathrm{method}\mathrm{is}\mathrm{paradox}.$$$$\mathrm{Actually}\mathrm{your}\mathrm{answer}\mathrm{suggest}\mathrm{AB}=0$$$$\mathrm{also},\mathrm{which}\mathrm{you}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{consider}.$$

Commented by Rasheed.Sindhi last updated on 26/May/18

$$\mathrm{A}+\mathrm{B}=10\wedge \mathrm{B}-\mathrm{A}=6\wedge \mathrm{AB}=0$$$$\mathrm{There}{\mathrm{aren}}^{\prime }\mathrm{t}\mathrm{any}\mathrm{values}\mathrm{which}\mathrm{satisfy}$$$$\mathrm{all}\mathrm{these}\mathrm{equations}\mathrm{simultaneously}.$$$$\mathrm{If}\mathrm{these}\mathrm{all}\mathrm{condions}\mathrm{were}\mathrm{satisfied}\mathrm{by}$$$$\mathrm{by}\mathrm{some}\mathrm{values}\mathrm{simultaneously},\mathrm{your}$$$$\mathrm{method}\mathrm{might}\mathrm{yeild}\mathrm{correct}\mathrm{result}.$$$$\mathrm{So}\mathrm{the}\mathrm{method}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{paradox}!$$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18

$$thankyou..ithoughtimadeahugemkstake..$$

Commented by behi83417@gmail.com last updated on 26/May/18

$$x^4+10x^3+6x-1=$$$$(x^2+\frac{\sqrt{5}-1}{2})(x^2+10x-\frac{\sqrt{5}+1}{2})$$$$whatdoyouthinkmrtanmay\mathrm{\&}mrRasheed?$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$\mathrm{What}\mathrm{method}\mathrm{did}\mathrm{you}\mathrm{use}\mathrm{to}\mathrm{factorize}?$$$$\mathrm{The}\mathrm{method}\mathrm{Mr}\mathrm{tanmay}\mathrm{used}\mathrm{is}\mathrm{failed}$$$$\mathrm{in}\mathrm{this}\mathrm{particular}\mathrm{case}.$$

Commented by $@ty@m last updated on 27/May/18

$$Infactitiswrongtoassumethat$$$$thequadraticfactorshave1and-1$$$$asconstantterms.$$$$Theproductoftwoirratioal$$$$numbersmayalsoyield1asin$$$$thepresentcase.$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$$\mathrm{Good}\mathrm{approach}\mathrm{sir},$$$$\mathrm{But}\left(ii\right)\mathrm{should}\mathrm{be},\mathrm{I}\mathrm{think}$$$$ab+ac+ad+bc+bd+cd=0$$$$\mathrm{instead}\mathrm{of}$$$$(a+b)(c+d)=0$$$$...Ifcoefficientofthehighesttermis$$$$unity:$$$$Thesumoftheproductsoftheroots,$$$$takentwoatatime,isequaltothe$$$$coefficientofthethirdterm.$$

Commented by Rasheed.Sindhi last updated on 27/May/18

$${\mathrm{t}}^4+{10\mathrm{t}}^3+6\mathrm{t}-1=0$$$$\mathrm{If}\mathrm{one}\mathrm{root}\mathrm{is}\sqrt{\frac{3}{5}}i$$$${\left(\sqrt{\frac{3}{5}}i\right)}^4+10{\left(\sqrt{\frac{3}{5}}i\right)}^3+6\left(\sqrt{\frac{3}{5}}i\right)-1=0$$$$\frac{9}{25}+10\left(\sqrt{\frac{3}{5}}i\right){\left(\sqrt{\frac{3}{5}}i\right)}^2+6\mathrm{i}\sqrt{\frac{3}{5}}-1=0$$$$\frac{9}{25}+10\times -\frac{3}{5}\times \sqrt{\frac{3}{5}}i+6\mathrm{i}\sqrt{\frac{3}{5}}-1=0$$$$\frac{9}{25}-6\mathrm{i}\sqrt{\frac{3}{5}}+6\mathrm{i}\sqrt{\frac{3}{5}}-1=0$$$$\frac{9}{25}-1\ne 0$$

Commented by $@ty@m last updated on 28/May/18

$$Youareright.$$$$I^{\prime }lltryagain.$$$$Thanksforcorrectingme.$$

Commented by MJS last updated on 29/May/18

$$(x^2+\frac{\sqrt{5}-1}{2})(x^2+10x-\frac{\sqrt{5}+1}{2})=$$$$=x^4+10x^3-x^2+5(\sqrt{5}-1)x-1$$$$\mathrm{so}\mathrm{this}\mathrm{is}\mathrm{wrong}$$

Answered by Rasheed.Sindhi last updated on 28/May/18

$$.....$$$$........$$$${\mathrm{t}}^4+{10\mathrm{t}}^3+6\mathrm{t}-1=0\left(\mathrm{Drived}\mathrm{by}\mathrm{Mr}\mathrm{tanmay}\right)$$$$\Rightarrow ({\mathrm{t}}^2+\mathrm{At}+\mathrm{B})({\mathrm{t}}^2+\mathrm{Ct}+\mathrm{D})=0\left(\mathrm{Say}\right)$$$$\mathrm{Where}\mathrm{BD}=-1$$$${\mathrm{t}}^4+{\mathrm{Ct}}^3+{\mathrm{Dt}}^2$$$$+{\mathrm{At}}^3+{\mathrm{ACt}}^2+\mathrm{ADt}$$$$+{\mathrm{Bt}}^2+\mathrm{BCt}+\mathrm{BD}(=-1)=0$$$$------------------$$$${\mathrm{t}}^4+(\mathrm{A}+\mathrm{C}){\mathrm{t}}^3+(\mathrm{AC}+\mathrm{B}+\mathrm{D}){\mathrm{t}}^2$$$$+(\mathrm{AD}+\mathrm{BC})\mathrm{t}-1=0$$$$\mathrm{A}+\mathrm{C}=10,\mathrm{AC}+\mathrm{B}+\mathrm{D}=0,\mathrm{AD}+\mathrm{BC}=6$$$$(\mathrm{BD}=-1$$$$\mathrm{A}+\mathrm{C}=10$$$$\mathrm{AC}+\mathrm{B}+\mathrm{D}=0$$$$\mathrm{AD}+\mathrm{BC}=6$$$$\mathrm{are}\mathrm{the}\mathrm{same}\mathrm{equations}\mathrm{as}\mathrm{in}\mathrm{an}\mathrm{other}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{It}\mathrm{seems}\mathrm{that}\mathrm{Mr}\mathrm{behi}\mathrm{has}\mathrm{applied}$$$$\mathrm{above}\mathrm{approach}\mathrm{and}\mathrm{encounterd}$$$$\mathrm{the}\mathrm{above}\mathrm{set}\mathrm{of}\mathrm{equations}\mathrm{and}\mathrm{he}$$$$\mathrm{posted}\mathrm{it}\mathrm{as}\mathrm{an}\mathrm{other}\mathrm{question}.$$$$\mathrm{Am}\mathrm{I}\mathrm{right}\mathrm{Mr}\mathrm{Abu}\mathrm{Behi}!:)$$$$$$$$....$$

Commented by behi83417@gmail.com last updated on 28/May/18

$$mrRasheed!youareright.$$$$butihavenotsolutionforthisand$$$$Iamwaitingformyfriendssuchas$$$$youandmrtanmayandmrsatyam..!$$

Commented by Rasheed.Sindhi last updated on 28/May/18

$$\mathrm{Mr}\mathrm{Abu}\mathrm{Behi}$$$$\mathrm{You}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{solution}\mathrm{but}\mathrm{at}\mathrm{least}$$$$\mathrm{you}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{factorize}\mathrm{the}\mathrm{following}:$$$$x^4+10x^3+6x-1$$$$=(x^2+\frac{\sqrt{5}-1}{2})(x^2+10x-\frac{\sqrt{5}+1}{2})$$$$---------------$$$$(x^2+\frac{\sqrt{5}-1}{2})(x^2+10x-\frac{\sqrt{5}+1}{2})=0$$$$\Rightarrow x^2=-\frac{\sqrt{5}-1}{2}<0\Rightarrow x\mathrm{is}\mathrm{not}\mathrm{real}$$$$\mathrm{Or}$$$$\Rightarrow x=\frac{-10\pm \sqrt{100+2\sqrt{5}+2}}{2}$$$$\tan \left(\frac{\theta }{2}\right)=\frac{-10\pm \sqrt{100+2\sqrt{5}+2}}{2}$$$$\frac{\theta }{2}={\tan }^{-1}\left(\frac{-10\pm \sqrt{100+2\sqrt{5}+2}}{2}\right)$$$$\theta ={2\tan }^{-1}\left(\frac{-10\pm \sqrt{100+2\sqrt{5}+2}}{2}\right)$$$$\theta =18.10°,-168.76°\left(\mathrm{out}\mathrm{of}\mathrm{permited}\mathrm{range}\right)$$$$\theta =18.10°=0.31588\mathrm{rad}$$

Commented by behi83417@gmail.com last updated on 31/May/18

$$dearmrRasheed!$$$$thisisthefinalanswerofourequation.$$$$itjusthaveonlyonerootin(0,\frac{\pi }{2})and$$$$itis:\theta =18.15\left(DEG\right)or0.32\left(RAD\right)$$

Answered by Rasheed.Sindhi last updated on 28/May/18

$${(1+\sin \theta )}^2={\left(\frac{4\sin \theta }{\cos \theta }\right)}^2$$$$1+2\sin \theta +{\sin }^2\theta =\frac{{16\sin }^2\theta }{{\cos }^2\theta }$$$$1+2\sin \theta +{\sin }^2\theta =\frac{{16\sin }^2\theta }{1-{\sin }^2\theta }$$$$1+2\sin \theta +{\sin }^2\theta -{\sin }^2\theta -{2\sin }^3\theta -{\sin }^4\theta$$$$-{16\sin }^2\theta =0$$$$-{\sin }^4\theta -{2\sin }^3\theta -{16\sin }^2\theta +2\sin \theta +1=0$$$${\sin }^4\theta +{2\sin }^3\theta +{16\sin }^2\theta -2\sin \theta -1=0$$$$\sin \theta =\mathrm{x}$$$${\mathrm{x}}^4+{2\mathrm{x}}^3+{16\mathrm{x}}^2-2\mathrm{x}-1=0$$$$\mathrm{Let}{\mathrm{x}}^4+{2\mathrm{x}}^3+{16\mathrm{x}}^2-2\mathrm{x}-1$$$$=({\mathrm{x}}^2+\mathrm{Ax}+\mathrm{B})({\mathrm{x}}^2+\mathrm{Cx}+\mathrm{D})$$$$={\mathrm{x}}^4+{\mathrm{Ax}}^3+{\mathrm{Bx}}^2$$$$+{\mathrm{Cx}}^3+{\mathrm{ACx}}^2+\mathrm{BCx}$$$$+{\mathrm{Dx}}^2+\mathrm{ADx}+\mathrm{BD}$$$$------------------$$$$\mathrm{A}+\mathrm{C}=2$$$$\mathrm{B}+\mathrm{AC}+\mathrm{D}=16$$$$\mathrm{AD}+\mathrm{BC}=-2$$$$\mathrm{BD}=-1$$$$\mathrm{An}\mathrm{other}\mathrm{set}\mathrm{of}\mathrm{equations}\left(\mathrm{same}\mathrm{composition}\right)$$$$.....$$

Commented by behi83417@gmail.com last updated on 28/May/18

$$mrRasheed!thankyouverymuch.$$$$buthowtosolvethisset?newproblem!$$

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