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Question Number 191873 by cherokeesay last updated on 02/May/23

Answered by mehdee42 last updated on 03/May/23

let : ∠RPS=p_1 &∠ QPS=p_2 p_1 =45−x , p_2 =135−3x ((PS)/(QS))=((sin3x)/(sinp_2 ))=((sinx)/(sinp_1 )) ⇒((sin3x)/(sin(135−3x)))=((sinx)/(sin(45−x))) ⇒sin3x(cosx−sinx)=sinx(cos3x+sin3x) sin3xcosx−sinxcos3x=2sinxsin3x sin2x=2sinxsin3x⇒sinx(sin3x−cosx)=0 in the triangle ; sinx≠0 ⇒sin3x=cosx=sin(90−x) ⇒3x=360k+90−x⇒x=22.5 ✓ & 3x=360k+90+x⇒x=45 #

$${let}\:\::\:\angle{RPS}={p}_{\mathrm{1}} \:\&\angle\:{QPS}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} =\mathrm{45}−{x}\:\:\:,\:\:{p}_{\mathrm{2}} =\mathrm{135}−\mathrm{3}{x} \\ $$$$\frac{{PS}}{{QS}}=\frac{{sin}\mathrm{3}{x}}{{sinp}_{\mathrm{2}} }=\frac{{sinx}}{{sinp}_{\mathrm{1}} }\:\Rightarrow\frac{{sin}\mathrm{3}{x}}{{sin}\left(\mathrm{135}−\mathrm{3}{x}\right)}=\frac{{sinx}}{{sin}\left(\mathrm{45}−{x}\right)} \\ $$$$\Rightarrow{sin}\mathrm{3}{x}\left({cosx}−{sinx}\right)={sinx}\left({cos}\mathrm{3}{x}+{sin}\mathrm{3}{x}\right) \\ $$$${sin}\mathrm{3}{xcosx}−{sinxcos}\mathrm{3}{x}=\mathrm{2}{sinxsin}\mathrm{3}{x} \\ $$$${sin}\mathrm{2}{x}=\mathrm{2}{sinxsin}\mathrm{3}{x}\Rightarrow{sinx}\left({sin}\mathrm{3}{x}−{cosx}\right)=\mathrm{0} \\ $$$${in}\:{the}\:{triangle}\:;\:{sinx}\neq\mathrm{0} \\ $$$$\Rightarrow{sin}\mathrm{3}{x}={cosx}={sin}\left(\mathrm{90}−{x}\right) \\ $$$$\Rightarrow\mathrm{3}{x}=\mathrm{360}{k}+\mathrm{90}−{x}\Rightarrow{x}=\mathrm{22}.\mathrm{5}\:\checkmark \\ $$$$\&\:\mathrm{3}{x}=\mathrm{360}{k}+\mathrm{90}+{x}\Rightarrow{x}=\mathrm{45}\:# \\ $$$$ \\ $$

Commented by mehdee42 last updated on 03/May/23

this triangle is vertical at the vertex P.

$${this}\:{triangle}\:{is}\:{vertical}\:{at}\:{the}\:{vertex}\:{P}. \\ $$

Answered by a.lgnaoui last updated on 03/May/23

△PQS ((sin 45)/(PQ))=((sin 3x)/(PS))=((sin (3x+45))/(QS)) ⇒PS=(((√2) PQsin 3x)/( 1))=((QSsin 3x)/(sin (3x+45))) PQ=((QS)/( (√2)sin (3x+45))) △PRS ∡PSR=((3𝛑)/4) ((sin x)/(PS))=((sin 135)/(PR))=((sin (45−x)))/(SR)) ⇒ PS=((PRsin x)/(sin 135)) •PQR ((sin x)/(PQ=))=((sin 3x)/(PR))=((sin (𝛑−4x))/(QS+SR)) (3) (1) ((sin 45)/(PQ))=((sin( 3x+45))/(QS)) QS=((PQsin (3x+45))/(sin 45)) (2) SR=(((45−X)×PR)/(sin 135)) QS+SR=((2PQsin (3X+45))/( (√2)))+((sin (45−x)×PR)/(sin 135)) (3)⇒PQ=((PRsin x)/(sin 3x)) ⇒[((sin x×sin (3x+45))/( (√2)))+((sin (45−x))/(sin 135))]PR=QS+SR (3)⇒((sin3 x)/(PR))=[(((√2) sin 135sin 4x)/( (√2) sin (45−x)+sin 135×sin xsin (3x+45)PR))] sin 3x((√2) sin (45−x)+sin 135sin( 3x+45)sin x]= (√(2 )) sin 135×sin 4x to continous........... sin 3x[(√2) (((√2)/2)cos x−((√2)/2)sin x)+((√2)/2)sin x(((√2)/2)sin 3x+((√2)/2)cos 3x) =sin 4x sin 3x(cos x−sin x)+((sin x(sin 3x+cos 3x))/2)) sin 4x=sin (3x+x)= sin 3xcos x+cos 3xsin x=sin 3x(cos x+((sin x)/(tan 3x))) ⇒cos x+((sin x)/(tan 3x))=cos x−sin x+(1/2)(sin x+((sin x)/(tan 3x))) ((sin x)/(tan 3x))=((sin x)/(2tan 3x))−(1/2)sin x ⇒(1/(tan 3x))=(1/(2tan 3x))−(1/2) (1/(2tang3x))=−(1/2)⇒ (1/(tang3x))=− 1 tang 3x=tang(−(𝛑/4)+k𝛑) 3x≠(π/2)+kπ x≠kπ x=((−𝛑)/(12))+((k𝛑)/3) (k∈Z) x={−(𝛑/(12));(π/4);((3π)/4);((5π)/4);((11π)/(12))} retenus {(π/4),((3π)/4)((11π)/(12))}

$$\bigtriangleup\boldsymbol{\mathrm{PQS}}\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{45}}{\boldsymbol{\mathrm{PQ}}}=\frac{\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{PS}}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)}{\boldsymbol{\mathrm{QS}}} \\ $$$$\Rightarrow\mathrm{PS}=\frac{\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{PQ}}\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}}{\:\mathrm{1}}=\frac{\boldsymbol{\mathrm{QS}}\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{PQ}}=\frac{\boldsymbol{\mathrm{QS}}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)} \\ $$$$\: \\ $$$$\bigtriangleup\mathrm{PRS}\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{PSR}}=\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$\:\:\:\:\frac{\mathrm{sin}\:\mathrm{x}}{\boldsymbol{\mathrm{PS}}}=\frac{\mathrm{sin}\:\mathrm{135}}{\mathrm{PR}}=\frac{\left.\mathrm{sin}\:\left(\mathrm{45}−\boldsymbol{\mathrm{x}}\right)\right)}{\boldsymbol{\mathrm{SR}}}\:\:\:\:\:\: \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{PS}}=\frac{\boldsymbol{\mathrm{PR}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:\mathrm{135}} \\ $$$$\bullet\boldsymbol{\mathrm{PQR}}\:\:\:\:\frac{\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{PQ}}=}=\frac{\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{PR}}}=\frac{\mathrm{sin}\:\left(\boldsymbol{\pi}−\mathrm{4x}\right)}{\boldsymbol{\mathrm{QS}}+\boldsymbol{\mathrm{SR}}}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)\:\:\:\frac{\mathrm{sin}\:\mathrm{45}}{\boldsymbol{\mathrm{PQ}}}=\frac{\mathrm{sin}\left(\:\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)}{\boldsymbol{\mathrm{QS}}} \\ $$$$\boldsymbol{\mathrm{QS}}=\frac{\boldsymbol{\mathrm{PQ}}\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)}{\mathrm{sin}\:\mathrm{45}} \\ $$$$\left(\mathrm{2}\right)\:\:\:\boldsymbol{\mathrm{SR}}=\frac{\left(\mathrm{45}−\boldsymbol{\mathrm{X}}\right)×\boldsymbol{\mathrm{PR}}}{\mathrm{sin}\:\mathrm{135}} \\ $$$$\boldsymbol{\mathrm{QS}}+\boldsymbol{\mathrm{SR}}=\frac{\mathrm{2}\boldsymbol{\mathrm{PQ}}\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{X}}+\mathrm{45}\right)}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{sin}\:\left(\mathrm{45}−\boldsymbol{\mathrm{x}}\right)×\boldsymbol{\mathrm{PR}}}{\mathrm{sin}\:\mathrm{135}} \\ $$$$\left(\mathrm{3}\right)\Rightarrow\boldsymbol{\mathrm{PQ}}=\frac{\boldsymbol{\mathrm{PR}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}}{\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}} \\ $$$$\Rightarrow\left[\frac{\mathrm{sin}\:\boldsymbol{\mathrm{x}}×\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{sin}\:\left(\mathrm{45}−\boldsymbol{\mathrm{x}}\right)}{\mathrm{sin}\:\mathrm{135}}\right]\boldsymbol{\mathrm{PR}}=\boldsymbol{\mathrm{QS}}+\boldsymbol{\mathrm{SR}} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\Rightarrow\frac{\mathrm{sin3}\:\mathrm{x}}{\boldsymbol{\mathrm{PR}}}=\left[\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{135sin}\:\mathrm{4}\boldsymbol{\mathrm{x}}}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{45}−\boldsymbol{\mathrm{x}}\right)+\mathrm{sin}\:\mathrm{135}×\mathrm{sin}\:\boldsymbol{\mathrm{x}}\mathrm{sin}\:\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)\boldsymbol{\mathrm{PR}}}\right] \\ $$$$\mathrm{sin}\:\mathrm{3}\boldsymbol{\mathrm{x}}\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{45}−\boldsymbol{\mathrm{x}}\right)+\mathrm{sin}\:\mathrm{135sin}\left(\:\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{45}\right)\mathrm{sin}\:\mathrm{x}\right]= \\ $$$$\sqrt{\mathrm{2}\:}\:\mathrm{sin}\:\mathrm{135}×\mathrm{sin}\:\mathrm{4}\boldsymbol{\mathrm{x}} \\ $$$$\:\:\mathrm{to}\:\mathrm{continous}........... \\ $$$$\mathrm{sin}\:\mathrm{3x}\left[\sqrt{\mathrm{2}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\mathrm{3x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos}\:\mathrm{3x}\right)\right. \\ $$$$=\mathrm{sin}\:\mathrm{4x} \\ $$$$\left.\mathrm{sin}\:\mathrm{3x}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)+\frac{\mathrm{sin}\:\mathrm{x}\left(\mathrm{sin}\:\mathrm{3x}+\mathrm{cos}\:\mathrm{3x}\right)}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:\mathrm{4x}=\mathrm{sin}\:\left(\mathrm{3x}+\mathrm{x}\right)= \\ $$$$\mathrm{sin}\:\mathrm{3xcos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{3xsin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{3x}\left(\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{3x}}\right) \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{3x}}=\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{3x}}\right) \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{3x}}=\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2tan}\:\mathrm{3x}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{3}\boldsymbol{\mathrm{x}}}=\frac{\mathrm{1}}{\mathrm{2tan}\:\mathrm{3}\boldsymbol{\mathrm{x}}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{tang}}\mathrm{3}\boldsymbol{\mathrm{x}}}=−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\:\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{tang}}\mathrm{3}\boldsymbol{\mathrm{x}}}=−\:\:\mathrm{1} \\ $$$$\:\:\:\boldsymbol{\mathrm{tang}}\:\mathrm{3}\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{tang}}\left(−\frac{\boldsymbol{\pi}}{\mathrm{4}}+\mathrm{k}\boldsymbol{\pi}\right)\:\:\: \\ $$$$\mathrm{3x}\neq\frac{\pi}{\mathrm{2}}+\mathrm{k}\pi\:\:\mathrm{x}\neq\mathrm{k}\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{−\boldsymbol{\pi}}{\mathrm{12}}+\frac{\mathrm{k}\boldsymbol{\pi}}{\mathrm{3}}\:\:\:\:\:\:\:\left(\mathrm{k}\in\mathbb{Z}\right) \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}=\left\{−\frac{\boldsymbol{\pi}}{\mathrm{12}};\frac{\pi}{\mathrm{4}};\frac{\mathrm{3}\pi}{\mathrm{4}};\frac{\mathrm{5}\pi}{\mathrm{4}};\frac{\mathrm{11}\pi}{\mathrm{12}}\right\} \\ $$$$\mathrm{retenus}\:\:\:\left\{\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}}\frac{\mathrm{11}\pi}{\mathrm{12}}\right\} \\ $$

Commented by mr W last updated on 03/May/23

a triangle with negative angles? how does the triangle look like with x=−(π/(12)) or with x=((5π)/4)?

$${a}\:{triangle}\:{with}\:{negative}\:{angles}? \\ $$$${how}\:{does}\:{the}\:{triangle}\:{look}\:{like}\:{with} \\ $$$${x}=−\frac{\pi}{\mathrm{12}}\:{or}\:{with}\:{x}=\frac{\mathrm{5}\pi}{\mathrm{4}}? \\ $$

Commented by maths_plus last updated on 03/May/23

waoh !

$$\mathrm{waoh}\:! \\ $$

Commented by a.lgnaoui last updated on 03/May/23

for forme triangle given the solution conforme are only angles: (π/4),((3π)/4), ((5π)/4); ((11π)/(12)).

$$\mathrm{for}\:\mathrm{forme}\:\mathrm{triangle}\:\mathrm{given}\:\mathrm{the}\:\mathrm{solution}\:\:\mathrm{conforme}\:\mathrm{are} \\ $$$$\:\mathrm{only}\:\mathrm{angles}:\:\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}},\:\frac{\mathrm{5}\pi}{\mathrm{4}};\:\frac{\mathrm{11}\pi}{\mathrm{12}}. \\ $$

Commented by mr W last updated on 03/May/23

then please draw a triangle with any of these values, e.g. x=((11π)/(12)).

$${then}\:{please}\:{draw}\:{a}\:{triangle}\:{with} \\ $$$${any}\:{of}\:{these}\:{values},\:{e}.{g}.\:\:{x}=\frac{\mathrm{11}\pi}{\mathrm{12}}. \\ $$

Commented by a.lgnaoui last updated on 03/May/23

Commented by a.lgnaoui last updated on 03/May/23

((5π)/4): not valide

$$\frac{\mathrm{5}\pi}{\mathrm{4}}:\:\:\:\mathrm{not}\:\mathrm{valide} \\ $$

Commented by mr W last updated on 03/May/23

x=((11π)/(12))=165° ! 3x=495° !

$${x}=\frac{\mathrm{11}\pi}{\mathrm{12}}=\mathrm{165}°\:\:! \\ $$$$\mathrm{3}{x}=\mathrm{495}°\:! \\ $$

Commented by a.lgnaoui last updated on 03/May/23

495−360=135=((3π)/4)mod(2π)

$$\mathrm{495}−\mathrm{360}=\mathrm{135}=\frac{\mathrm{3}\pi}{\mathrm{4}}\mathrm{mod}\left(\mathrm{2}\pi\right) \\ $$

Commented by a.lgnaoui last updated on 03/May/23

⇒495⇔135°[2π] we can retire it from the solution becsuse its in the contre−sens of angle as in the guven trisngle Remsrque: if the triangle is given in the contraire sens like (joint picture) x take value(−x)

$$\Rightarrow\mathrm{495}\Leftrightarrow\mathrm{135}°\left[\mathrm{2}\pi\right] \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{retire}\:\mathrm{it}\:\mathrm{from}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{becsuse}\:\mathrm{its}\:\mathrm{in}\:\mathrm{the}\:\mathrm{contre}−\mathrm{sens}\:\mathrm{of}\:\mathrm{angle} \\ $$$$\mathrm{as}\:\mathrm{in}\:\mathrm{the}\:\mathrm{guven}\:\mathrm{trisngle}\: \\ $$$$\boldsymbol{\mathrm{Remsrque}}:\:\mathrm{if}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\: \\ $$$$\mathrm{the}\:\mathrm{contraire}\:\mathrm{sens}\:\mathrm{like}\:\left(\mathrm{joint}\:\mathrm{picture}\right) \\ $$$$\mathrm{x}\:\mathrm{take}\:\mathrm{value}\left(−\mathrm{x}\right) \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 03/May/23

Commented by mr W last updated on 04/May/23

then draw exactly a triangle with your solution x=(π/4) !

$${then}\:{draw}\:{exactly}\:{a}\:{triangle}\:{with} \\ $$$${your}\:{solution}\:{x}=\frac{\pi}{\mathrm{4}}\:! \\ $$

Answered by ajfour last updated on 04/May/23

QP y=(((3m−m^3 )/(1−3m^2 )))(x+1) y=−x y=m(1−x) ⇒ m−mx+x=0 x=(m/(m−1)) (m/(1−m))=m(((3−m^2 )/(1−3m^2 )))((m/(m−1))+1) ⇒ (3−m^2 )(2m−1)=3m^2 −1 ⇒ 2m^3 +2m^2 −6m+2=0 ⇒ m^3 +m^2 −3m+1=0 x=tan^(−1) m

$${QP} \\ $$$${y}=\left(\frac{\mathrm{3}{m}−{m}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{m}^{\mathrm{2}} }\right)\left({x}+\mathrm{1}\right) \\ $$$${y}=−{x} \\ $$$${y}={m}\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow\:\:{m}−{mx}+{x}=\mathrm{0} \\ $$$${x}=\frac{{m}}{{m}−\mathrm{1}} \\ $$$$\frac{{m}}{\mathrm{1}−{m}}={m}\left(\frac{\mathrm{3}−{m}^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}{m}^{\mathrm{2}} }\right)\left(\frac{{m}}{{m}−\mathrm{1}}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\left(\mathrm{3}−{m}^{\mathrm{2}} \right)\left(\mathrm{2}{m}−\mathrm{1}\right)=\mathrm{3}{m}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{2}{m}^{\mathrm{3}} +\mathrm{2}{m}^{\mathrm{2}} −\mathrm{6}{m}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{m}^{\mathrm{3}} +{m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} {m} \\ $$

Answered by mr W last updated on 04/May/23

say QS=SR=1 PQ=((sin 45°)/(sin (45°+3x))) PR=((sin 45°)/(sin (45°−x))) ((PQ)/(PR))=((sin x)/(sin 3x)) ((sin 45°)/(sin (45°+3x)))×((sin (45−x))/(sin 45°))=((sin x)/(sin 3x)) ((cos x−sin x)/(cos 3x+sin 3x))=((sin x)/(sin 3x)) ((1−tan x)/(1+tan 3x))=((tan x)/(tan 3x)) with t=tan x ((1−t)/(1+((3t−t^3 )/(1−3t^2 ))))=(t/((3t−t^3 )/(1−3t^2 ))) t^3 +t^2 −3t+1=0 t^3 −t^2 +2t^2 −2t−t+1=0 (t−1)(t^2 +2t−1)=0 ⇒t=1 ⇒ rejected! ⇒t=−1−(√2) ⇒rejected! ⇒t=−1+(√2) ⇒x=tan^(−1) ((√2)−1)=22.5°

$${say}\:{QS}={SR}=\mathrm{1} \\ $$$${PQ}=\frac{\mathrm{sin}\:\mathrm{45}°}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{3}{x}\right)} \\ $$$${PR}=\frac{\mathrm{sin}\:\mathrm{45}°}{\mathrm{sin}\:\left(\mathrm{45}°−{x}\right)} \\ $$$$\frac{{PQ}}{{PR}}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{45}°}{\mathrm{sin}\:\left(\mathrm{45}°+\mathrm{3}{x}\right)}×\frac{\mathrm{sin}\:\left(\mathrm{45}−{x}\right)}{\mathrm{sin}\:\mathrm{45}°}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}} \\ $$$$\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{3}{x}+\mathrm{sin}\:\mathrm{3}{x}}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}} \\ $$$$\frac{\mathrm{1}−\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:\mathrm{3}{x}}=\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:\mathrm{3}{x}} \\ $$$${with}\:{t}=\mathrm{tan}\:{x} \\ $$$$\frac{\mathrm{1}−{t}}{\mathrm{1}+\frac{\mathrm{3}{t}−{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} }}=\frac{{t}}{\frac{\mathrm{3}{t}−{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} }} \\ $$$${t}^{\mathrm{3}} +{t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}−{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1}\:\Rightarrow\:{rejected}! \\ $$$$\Rightarrow{t}=−\mathrm{1}−\sqrt{\mathrm{2}}\:\Rightarrow{rejected}! \\ $$$$\Rightarrow{t}=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{22}.\mathrm{5}° \\ $$

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