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Question Number 185424 by ajfour last updated on 21/Jan/23

Commented by Frix last updated on 21/Jan/23

$$\mathrm{Impossible}.$$$$\tan \theta =\frac{(2(x-1)+\sqrt{x+2})\sqrt{x}}{4x-1}$$

Commented by MJS_new last updated on 22/Jan/23

$$\mathrm{possible}!$$$$k=\tan \theta$$$$k=\frac{(2(x-1)+\sqrt{x+2})\sqrt{x}}{4x-1}$$$$x=y^2\wedge y⩾0$$$$k=\frac{y(2(y^2-1)+\sqrt{y^2+2})}{4y^2-1}$$$$y=\frac{(z^2-1)\sqrt{2}}{2z}\wedge z⩾0$$$$k=\frac{(z^2-1)(\sqrt{2}{z}^2+4z+\sqrt{2})}{2z(2z^2+3\sqrt{2}z+2)}$$$$z^4-2\sqrt{2}(k-1)z^3-6{kz}^2-2\sqrt{2}(k+1)z-1=0$$$$\mathrm{the}\mathrm{usual}\mathrm{path}\mathrm{to}\mathrm{solve}\mathrm{a}\mathrm{quartic}\mathrm{leads}\mathrm{to}$$$$\mathrm{these}\mathrm{square}\mathrm{factors}:$$$$z^2-\sqrt{2}(k-1+\sqrt{k^2+1})z-k-\sqrt{k^2+1}=0$$$$z^2-\sqrt{2}(k-1-\sqrt{k^2+1})z-k+\sqrt{k^2+1}=0$$$$\mathrm{now}\mathrm{solve}\mathrm{these}\mathrm{and}\mathrm{insert}\mathrm{backwards}:$$$$y=\frac{(z^2-1)\sqrt{2}}{2z}\wedge z⩾0$$$$x=y^2\wedge y⩾0$$$$k=\tan \theta$$

Commented by Frix last updated on 21/Jan/23

$$\mathrm{Great}!$$

Commented by ajfour last updated on 21/Jan/23

$$ohitcreatedsomuchmess,iam$$$$soglad!$$

Commented by MJS_new last updated on 21/Jan/23

$$\mathrm{have}\mathrm{you}\mathrm{found}\mathrm{another}\mathrm{solution}?$$$$\mathrm{I}\mathrm{first}\mathrm{tried}\mathrm{by}\mathrm{squaring}\mathrm{\&}\mathrm{transforming}2$$$$\mathrm{times}\mathrm{but}\mathrm{this}\mathrm{leads}\mathrm{to}{6}^{\mathrm{th}}\mathrm{grade}\mathrm{polynom}...$$

Answered by mr W last updated on 21/Jan/23

$$\sqrt{{(x+1)}^2-{(x-1)}^2}=\tan \frac{\theta }{2}+\frac{x}{\tan \theta }$$$$\frac{x}{\tan \theta }-2\sqrt{x}+\tan \frac{\theta }{2}=0$$$$\Rightarrow \sqrt{x}=\tan \theta (1+\sqrt{1-\frac{\tan \frac{\theta }{2}}{\tan \theta }})$$$$\Rightarrow x={\tan }^2\theta {(1+\sqrt{1-\frac{\tan \frac{\theta }{2}}{\tan \theta }})}^2$$

Commented by MJS_new last updated on 21/Jan/23

$$\mathrm{nice}!$$$$\mathrm{I}\mathrm{only}\mathrm{tried}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{equation}\mathrm{Sir}\mathrm{Frix}$$$$\mathrm{had}\mathrm{posted}.\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{more}\mathrm{compact}$$

Commented by MJS_new last updated on 21/Jan/23

$$\mathrm{your}\mathrm{result}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}$$$$x=\frac{(1-\cos \theta )(2+\cos \theta +2\sqrt{1+\cos \theta })}{{\cos }^2\theta }$$$$\iff$$$$\cos \theta =\frac{-x+1+2x\sqrt{x+2}}{{(x+1)}^2}$$$$\Rightarrow$$$$\tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }=\frac{(2(x-1)+\sqrt{x+2})\sqrt{x}}{4x-1}$$$$\mathrm{which}\mathrm{is}\mathrm{Sir}{\mathrm{Frix}}^{\prime }\mathrm{s}\mathrm{result}$$

Commented by Ar Brandon last updated on 22/Jan/23

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Commented by Rasheed.Sindhi last updated on 22/Jan/23

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Commented by Spillover last updated on 23/Jan/23

$$whyareyouloughing?$$

Commented by MJS_new last updated on 23/Jan/23

$$\mathrm{because}\mathrm{they}\mathrm{are}\mathrm{sure}\mathrm{that}$$$$\mathrm{lemma}:\mathrm{MJS}\mathrm{\_}\mathrm{new}=\mathrm{Frix}$$$$\mathrm{is}\mathrm{true}.{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}\mathrm{for}\mathrm{which}\mathrm{theorem}\mathrm{this}$$$$\mathrm{is}\mathrm{needed}\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{sure}\mathrm{it}\mathrm{is}\mathrm{wrong}\mathrm{although}$$$$\mathrm{I}\mathrm{cannot}\mathrm{prove}\mathrm{this}\mathrm{within}\mathrm{this}\mathrm{vector}\mathrm{space}$$$${\mathrm{we}}^{\prime }\mathrm{re}\mathrm{in}.$$

Commented by Frix last updated on 23/Jan/23

$$\mathrm{Anyway},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{better}\mathrm{to}\mathrm{laugh}\mathrm{than}\mathrm{it}\mathrm{is}\mathrm{to}\mathrm{cry}...$$

Commented by Ar Brandon last updated on 23/Jan/23

Keep on entertaining us Sir MJFrix ��

Commented by Rasheed.Sindhi last updated on 23/Jan/23

$$\mathrm{MJS}\mathrm{\_}\mathrm{new}=\mathrm{Frix}isitselfatheorem,$$$$Ithink.$$$$(Pleasesir\mathrm{MJS}/\mathrm{Frix}execuseus$$$$forourcuriosity.)$$

Commented by MJS_new last updated on 23/Jan/23

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{ok}.$$$$\mathrm{but}\mathrm{why}\mathrm{would}\mathrm{I}\mathrm{run}2\mathrm{accounts}?\mathrm{do}\mathrm{I}\mathrm{have}$$$$\mathrm{too}\mathrm{much}\mathrm{time}\mathrm{to}\mathrm{spend}?$$

Commented by Ar Brandon last updated on 23/Jan/23

Maybe her Majesty will be able to answer �� Good evening, sir MJFrix ��

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