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Question Number 82654 by jagoll last updated on 23/Feb/20
If sec x + tan x = 2+(√5) find sin x+ cos x ?
$$\boldsymbol{\mathrm{I}}\mathrm{f}\:\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{find}\:\mathrm{sin}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{x}\:? \\ $$
Commented by jagoll last updated on 23/Feb/20
Answered by mr W last updated on 23/Feb/20
((1+sin x)/(cos x))=2+(√5) cos x≠0 ⇒x≠2kπ±(π/2) (2+(√5))cos x−sin x=1 ((2+(√5))/( (√(1^2 +(2+(√5))^2 )))) cos x−(1/( (√(1^2 +(2+(√5))^2 )))) sin x=(1/( (√(1^2 +(2+(√5))^2 )))) cos α cos x−sin α sin x=sin α cos (x+α)=sin α=cos ((π/2)−α) ⇒x+α=2kπ±((π/2)−α) ⇒x=2kπ±((π/2)−α)−α= { ((2kπ+(π/2)−2α)),((2kπ−(π/2) ⇒not suitable)) :} ⇒x=2kπ+(π/2)−2α cos x+sin x=sin 2α+cos 2α =((2(2+(√5)))/(10+4(√5)))+(((2+(√5))^2 −1)/(10+4(√5))) =((3(2+(√5)))/(5+2(√5))) =((3(2+(√5)))/( (√5)(2+(√5)))) =((3(√5))/5)
$$\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{cos}\:{x}\neq\mathrm{0}\:\Rightarrow{x}\neq\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\mathrm{cos}\:{x}−\mathrm{sin}\:{x}=\mathrm{1} \\ $$$$\frac{\mathrm{2}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }}\:\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:{x}−\mathrm{sin}\:\alpha\:\mathrm{sin}\:{x}=\mathrm{sin}\:\alpha \\ $$$$\mathrm{cos}\:\left({x}+\alpha\right)=\mathrm{sin}\:\alpha=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow{x}+\alpha=\mathrm{2}{k}\pi\pm\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi\pm\left(\frac{\pi}{\mathrm{2}}−\alpha\right)−\alpha=\begin{cases}{\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha}\\{\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{2}}\:\Rightarrow{not}\:{suitable}}\end{cases} \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}=\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}}}+\frac{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\:\sqrt{\mathrm{5}}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$
Commented by jagoll last updated on 23/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 23/Feb/20
if possible one should always try without squaring to avoid that “fake” roots are added. with x=2kπ+(π/2)−2α i′m sure all true roots are included and no fake root is included.
$${if}\:{possible}\:{one}\:{should}\:{always}\:{try}\: \\ $$$${without}\:{squaring}\:{to}\:{avoid}\:{that}\:“{fake}'' \\ $$$${roots}\:{are}\:{added}.\:{with}\:{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$$${i}'{m}\:{sure}\:{all}\:{true}\:{roots}\:{are}\:{included}\:{and} \\ $$$${no}\:{fake}\:{root}\:{is}\:{included}.\: \\ $$
Answered by TANMAY PANACEA last updated on 23/Feb/20
(secx+tanx)(secx−tanx)=1 secx−tanx=(1/( (√5) +2))=(√5) −2 (secx+tanx)+(secx−tanx)=2(√5) secx=(√5) →cosx=(1/( (√5))) sinx=(2/( (√5)))→sinx+cosx=(3/( (√5)))=((3(√5))/5)
$$\left({secx}+{tanx}\right)\left({secx}−{tanx}\right)=\mathrm{1} \\ $$$${secx}−{tanx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:+\mathrm{2}}=\sqrt{\mathrm{5}}\:−\mathrm{2} \\ $$$$\left({secx}+{tanx}\right)+\left({secx}−{tanx}\right)=\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$${secx}=\sqrt{\mathrm{5}}\:\rightarrow{cosx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${sinx}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\rightarrow{sinx}+{cosx}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 23/Feb/20
nice solution!
$${nice}\:{solution}! \\ $$
Commented by TANMAY PANACEA last updated on 23/Feb/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by john santu last updated on 23/Feb/20
let : sec x − tan x = t (sec x+tan x)(sec x−tan x) = (2+(√5) )t sec^2 x−tan^2 x = (2+(√5))t tan^2 x+1−tan^2 x= (2+(√5)) t t = (1/(2+(√5)))= (√5) −2= (1/(cos x))−((sin x)/(cos x)) (1) from (1/(cos x))+((sin x)/(cos x)) = 2+(√5) (2) (1)+(2) ⇒ (2/(cos x)) = 2(√5) ⇒ cos x = (1/( (√5))) then sin x = (2/( (√5))) ⇒ sin x+cos x = (3/( (√5)))
$${let}\::\:\mathrm{sec}\:{x}\:−\:\mathrm{tan}\:{x}\:=\:{t}\: \\ $$$$\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)\:=\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\:\right){t} \\ $$$$\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{tan}\:^{\mathrm{2}} {x}\:=\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\right){t} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}=\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:{t} \\ $$$${t}\:=\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{5}}}=\:\sqrt{\mathrm{5}}\:−\mathrm{2}=\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:\left(\mathrm{1}\right) \\ $$$${from}\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\:=\:\mathrm{2}+\sqrt{\mathrm{5}}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{cos}\:{x}}\:=\:\mathrm{2}\sqrt{\mathrm{5}}\:\Rightarrow\:\mathrm{cos}\:{x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$${then}\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}} \\ $$
Commented by peter frank last updated on 23/Feb/20
thank you both
$${thank}\:{you}\:{both} \\ $$

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