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Question Number 69535 by naka3546 last updated on 24/Sep/19
Commented by naka3546 last updated on 24/Sep/19
without u
$${without}\:\:{u} \\ $$
Answered by MJS last updated on 24/Sep/19
atan t =k(1+sec t) ⇒ k+kcos t −asin t =0 t=2arctan u 2((k−au)/(u^2 +1))=0 ⇒ u=(k/a) ⇒ t=2arctan (k/a) ⇒ 2arctan (k/3) +2arctan (k/4) +2arctan (k/5) +2arctan (k/6) =2π arctan (k/3) +arctan (k/4) +arctan (k/5) +arctan (k/6) =π tan (arctan a +arctan b)=((a+b)/(1−ab)) tan (arctan a +arctan b +arctan c)= =((a+b+c−abc)/(1−(ab+ac+bc))) tan (arctan a +arctan b +arctan c +arctan d)= =((a+b+c+d−(abc+abd+acd+bcd))/(1+abcd−(ab+ac+ad+bc+bdd+cd))) ⇒ −((18k(k^2 −19))/(k^4 −119k^2 +360))=0 ⇒ k=±(√(19))∨k=0 but k=0 ⇒ w=x=y=z=0 k=−(√(19)) doesn′t fit the above equation ⇒ k=(√(19))
$${a}\mathrm{tan}\:{t}\:={k}\left(\mathrm{1}+\mathrm{sec}\:{t}\right) \\ $$$$\Rightarrow \\ $$$${k}+{k}\mathrm{cos}\:{t}\:−{a}\mathrm{sin}\:{t}\:=\mathrm{0} \\ $$$${t}=\mathrm{2arctan}\:{u} \\ $$$$\mathrm{2}\frac{{k}−{au}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0}\:\Rightarrow\:{u}=\frac{{k}}{{a}}\:\Rightarrow\:{t}=\mathrm{2arctan}\:\frac{{k}}{{a}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2arctan}\:\frac{{k}}{\mathrm{3}}\:+\mathrm{2arctan}\:\frac{{k}}{\mathrm{4}}\:+\mathrm{2arctan}\:\frac{{k}}{\mathrm{5}}\:+\mathrm{2arctan}\:\frac{{k}}{\mathrm{6}}\:=\mathrm{2}\pi \\ $$$$\mathrm{arctan}\:\frac{{k}}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{{k}}{\mathrm{4}}\:+\mathrm{arctan}\:\frac{{k}}{\mathrm{5}}\:+\mathrm{arctan}\:\frac{{k}}{\mathrm{6}}\:=\pi \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{arctan}\:{a}\:+\mathrm{arctan}\:{b}\right)=\frac{{a}+{b}}{\mathrm{1}−{ab}} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{arctan}\:{a}\:+\mathrm{arctan}\:{b}\:+\mathrm{arctan}\:{c}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{a}+{b}+{c}−{abc}}{\mathrm{1}−\left({ab}+{ac}+{bc}\right)} \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{arctan}\:{a}\:+\mathrm{arctan}\:{b}\:+\mathrm{arctan}\:{c}\:+\mathrm{arctan}\:{d}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{a}+{b}+{c}+{d}−\left({abc}+{abd}+{acd}+{bcd}\right)}{\mathrm{1}+{abcd}−\left({ab}+{ac}+{ad}+{bc}+{bdd}+{cd}\right)} \\ $$$$\Rightarrow \\ $$$$−\frac{\mathrm{18}{k}\left({k}^{\mathrm{2}} −\mathrm{19}\right)}{{k}^{\mathrm{4}} −\mathrm{119}{k}^{\mathrm{2}} +\mathrm{360}}=\mathrm{0}\:\Rightarrow\:{k}=\pm\sqrt{\mathrm{19}}\vee{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{but}\:{k}=\mathrm{0}\:\Rightarrow\:{w}={x}={y}={z}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{k}=−\sqrt{\mathrm{19}}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{above}\:\mathrm{equation} \\ $$$$\Rightarrow \\ $$$${k}=\sqrt{\mathrm{19}} \\ $$
Commented by mind is power last updated on 24/Sep/19
nice solution sir ,k=0 we can have π,π,0,0
$$\:{nice}\:{solution}\:{sir}\:,{k}=\mathrm{0}\:\:\:\:{we}\:{can}\:{have}\:\pi,\pi,\mathrm{0},\mathrm{0} \\ $$
Commented by MJS last updated on 24/Sep/19
true
$$\mathrm{true} \\ $$
Answered by mr W last updated on 24/Sep/19
sin w=(k/3)(1+cos w) sin^2 w=(k^2 /3^2 )(1+cos^2 w+2 cos w) ⇒(1+(3^2 /k^2 ))cos^2 w+2 cos w+(1−(3^2 /k^2 ))=0 ⇒cos w=((−1±(3^2 /k^2 ))/(1+(3^2 /k^2 )))= { ((−1 ⇒w=π)),(((3^2 −k^2 )/(3^2 +k^2 ))) :} similarly ⇒cos x= { ((−1 ⇒x=π)),(((4^2 −k^2 )/(4^2 +k^2 ))) :} ⇒cos y= { ((−1 ⇒y=π)),(((5^2 −k^2 )/(5^2 +k^2 ))) :} ⇒cos z= { ((−1 ⇒z=π)),(((6^2 −k^2 )/(6^2 +k^2 ))) :} .... k=±(√(19))=±4.3589 ???
$$\mathrm{sin}\:{w}=\frac{{k}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{cos}\:{w}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{w}=\frac{{k}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} }\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{w}+\mathrm{2}\:\mathrm{cos}\:{w}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)\mathrm{cos}^{\mathrm{2}} \:{w}+\mathrm{2}\:\mathrm{cos}\:{w}+\left(\mathrm{1}−\frac{\mathrm{3}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{w}=\frac{−\mathrm{1}\pm\frac{\mathrm{3}^{\mathrm{2}} }{{k}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{2}} }{{k}^{\mathrm{2}} }}=\begin{cases}{−\mathrm{1}\:\Rightarrow{w}=\pi}\\{\frac{\mathrm{3}^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\end{cases} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{cos}\:{x}=\begin{cases}{−\mathrm{1}\:\Rightarrow{x}=\pi}\\{\frac{\mathrm{4}^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow\mathrm{cos}\:{y}=\begin{cases}{−\mathrm{1}\:\Rightarrow{y}=\pi}\\{\frac{\mathrm{5}^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow\mathrm{cos}\:{z}=\begin{cases}{−\mathrm{1}\:\Rightarrow{z}=\pi}\\{\frac{\mathrm{6}^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} +{k}^{\mathrm{2}} }}\end{cases} \\ $$$$…. \\ $$$${k}=\pm\sqrt{\mathrm{19}}=\pm\mathrm{4}.\mathrm{3589}\:??? \\ $$
Commented by mind is power last updated on 24/Sep/19
nice sir please how how you put { for two ligns/?
$$\:\:\:{nice}\:{sir}\:{please}\:{how}\:{how}\:{you}\:{put}\:\:\left\{\:{for}\:{two}\:{ligns}/?\right. \\ $$$$\: \\ $$$$ \\ $$
Commented by mr W last updated on 24/Sep/19
just use the third button from left on the bottom of menu. example { ((line 1)),((line 2)) :}
$${just}\:{use}\:{the}\:{third}\:{button}\:{from}\:{left}\:{on} \\ $$$${the}\:{bottom}\:{of}\:{menu}. \\ $$$${example}\:\begin{cases}{{line}\:\mathrm{1}}\\{{line}\:\mathrm{2}}\end{cases} \\ $$
Commented by mind is power last updated on 24/Sep/19
thanx
$$\boldsymbol{{thanx}} \\ $$
Answered by mind is power last updated on 24/Sep/19
⇒3sin(w)=k(cos(w)+1) 4sin(x)=k(cos(x)+1) 5sin(y)=k(cos(y)+1) 6sin(z)=k(cos(z)+1) ⇔3tg((w/2))=4tg((x/2))=5tg((y/2))=6tg((z/2))=k tg(a)+tg(b)=((sin(a+b))/(cos(a)cos(b))) ⇒tg(((x+w)/2))+tg(((z+y)/2))=((sin(((x+y+z+w)/2)))/(cos(((x+y)/2))cos(((z+w)/2)))) tg(a+b)=((tg(a)+tg(b))/(1−tg(a)tg(b))) tg(((x+w)/2))+tg(((y+z)/2))=((tg((x/2))+tg((w/2)))/(1−tg((x/2))tg((w/2))))+((tg((y/2))+tg((z/2)))/(1−tg((y/2)).tg((z/2))))=0 we have tg((w/2))=(k/3) tg((x/2))=(k/4) , tg((y/2))=(k/5), tg((z/2))=(k/6) ⇒(((k/3)+(k/4))/(1−(k^2 /(12))))+(((k/5)+(k/6))/(1−(k^2 /(30))))=0 ⇒((7k)/(12−k^2 ))+((11k)/(30−k^2 ))=0 ⇒k(((7.(30−k^2 )+11(12−k^2 ))/((12−k^2 )(30−k^2 ))))=0 ⇒k=0 or 210−7k^2 +132−11k^2 =0⇒k^2 =((342)/(18))=19 ⇒k=+_− (√(19)) k=−(√(19))⇒(w/2) ,(x/2),(y/2),(x/2)>(π/2)⇒x+y+z+w>4π>2π k=(√(19)) i will study this later k∈{0,(√(19))}
$$\Rightarrow\mathrm{3}{sin}\left({w}\right)={k}\left({cos}\left({w}\right)+\mathrm{1}\right) \\ $$$$\mathrm{4}{sin}\left({x}\right)={k}\left({cos}\left({x}\right)+\mathrm{1}\right) \\ $$$$\mathrm{5}{sin}\left({y}\right)={k}\left({cos}\left({y}\right)+\mathrm{1}\right) \\ $$$$\mathrm{6}{sin}\left({z}\right)={k}\left({cos}\left({z}\right)+\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{3}{tg}\left(\frac{{w}}{\mathrm{2}}\right)=\mathrm{4}{tg}\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{5}{tg}\left(\frac{{y}}{\mathrm{2}}\right)=\mathrm{6}{tg}\left(\frac{{z}}{\mathrm{2}}\right)={k} \\ $$$${tg}\left({a}\right)+{tg}\left({b}\right)=\frac{{sin}\left({a}+{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)} \\ $$$$\Rightarrow{tg}\left(\frac{{x}+{w}}{\mathrm{2}}\right)+{tg}\left(\frac{{z}+{y}}{\mathrm{2}}\right)=\frac{{sin}\left(\frac{{x}+{y}+{z}+{w}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}+{y}}{\mathrm{2}}\right){cos}\left(\frac{{z}+{w}}{\mathrm{2}}\right)} \\ $$$${tg}\left({a}+{b}\right)=\frac{{tg}\left({a}\right)+{tg}\left({b}\right)}{\mathrm{1}−{tg}\left({a}\right){tg}\left({b}\right)} \\ $$$${tg}\left(\frac{{x}+{w}}{\mathrm{2}}\right)+{tg}\left(\frac{{y}+{z}}{\mathrm{2}}\right)=\frac{{tg}\left(\frac{{x}}{\mathrm{2}}\right)+{tg}\left(\frac{{w}}{\mathrm{2}}\right)}{\mathrm{1}−{tg}\left(\frac{{x}}{\mathrm{2}}\right){tg}\left(\frac{{w}}{\mathrm{2}}\right)}+\frac{{tg}\left(\frac{{y}}{\mathrm{2}}\right)+{tg}\left(\frac{{z}}{\mathrm{2}}\right)}{\mathrm{1}−{tg}\left(\frac{{y}}{\mathrm{2}}\right).{tg}\left(\frac{{z}}{\mathrm{2}}\right)}=\mathrm{0} \\ $$$${we}\:{have}\:{tg}\left(\frac{{w}}{\mathrm{2}}\right)=\frac{{k}}{\mathrm{3}} \\ $$$${tg}\left(\frac{{x}}{\mathrm{2}}\right)=\frac{{k}}{\mathrm{4}}\:\:,\:\:{tg}\left(\frac{{y}}{\mathrm{2}}\right)=\frac{{k}}{\mathrm{5}},\:\:\:{tg}\left(\frac{{z}}{\mathrm{2}}\right)=\frac{{k}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\frac{{k}}{\mathrm{3}}+\frac{{k}}{\mathrm{4}}}{\mathrm{1}−\frac{{k}^{\mathrm{2}} }{\mathrm{12}}}+\frac{\frac{{k}}{\mathrm{5}}+\frac{{k}}{\mathrm{6}}}{\mathrm{1}−\frac{{k}^{\mathrm{2}} }{\mathrm{30}}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{7}{k}}{\mathrm{12}−{k}^{\mathrm{2}} }+\frac{\mathrm{11}{k}}{\mathrm{30}−{k}^{\mathrm{2}} }=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{k}\left(\frac{\mathrm{7}.\left(\mathrm{30}−{k}^{\mathrm{2}} \right)+\mathrm{11}\left(\mathrm{12}−{k}^{\mathrm{2}} \right)}{\left(\mathrm{12}−{k}^{\mathrm{2}} \right)\left(\mathrm{30}−{k}^{\mathrm{2}} \right)}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\mathrm{0}\:{or}\:\:\:\mathrm{210}−\mathrm{7}{k}^{\mathrm{2}} +\mathrm{132}−\mathrm{11}{k}^{\mathrm{2}} =\mathrm{0}\Rightarrow{k}^{\mathrm{2}} =\frac{\mathrm{342}}{\mathrm{18}}=\mathrm{19} \\ $$$$ \\ $$$$\Rightarrow{k}=\underset{−} {+}\sqrt{\mathrm{19}} \\ $$$${k}=−\sqrt{\mathrm{19}}\Rightarrow\frac{{w}}{\mathrm{2}}\:,\frac{{x}}{\mathrm{2}},\frac{{y}}{\mathrm{2}},\frac{{x}}{\mathrm{2}}>\frac{\pi}{\mathrm{2}}\Rightarrow{x}+{y}+{z}+{w}>\mathrm{4}\pi>\mathrm{2}\pi\: \\ $$$${k}=\sqrt{\mathrm{19}}\:\:\:\:{i}\:{will}\:{study}\:{this}\:{later}\:{k}\in\left\{\mathrm{0},\sqrt{\mathrm{19}}\right\}\:\: \\ $$$$ \\ $$

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