Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

$${thank}\:{you}\:{both} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com