Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 175547 by daus last updated on 02/Sep/22

Answered by mr W last updated on 02/Sep/22

cos ((π/2)−x)=cos (x+(π/3))  ⇒2kπ+(π/2)−x=x+(π/3)  ⇒x=kπ+(π/(12))  in (−π,π): −((11π)/(12)), (π/(12))

$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−{x}={x}+\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}={k}\pi+\frac{\pi}{\mathrm{12}} \\ $$$${in}\:\left(−\pi,\pi\right):\:−\frac{\mathrm{11}\pi}{\mathrm{12}},\:\frac{\pi}{\mathrm{12}} \\ $$

Commented by Tawa11 last updated on 15/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by CElcedricjunior last updated on 02/Sep/22

sinx=cos(x+(𝛑/3)) for x∈]−π;𝛑]  or sinx=cos(−x+(𝛑/2))  =>cos(−x+(𝛑/2))=cos(x+(𝛑/3))  => { (((𝛑/2)−x=x+(𝛑/3)+2k𝛑)),(((π/2)−x=−x−(𝛑/3)+2k)) :}k∈Z  => { ((x=(𝛑/(12))+k𝛑)),() :}  S_(]−𝛑;𝛑]) ={−((11𝛑)/(12));(𝛑/(12))}               .......le celebre cedric junior.......

$$\left.\boldsymbol{{s}}\left.\boldsymbol{{inx}}=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)\:\boldsymbol{{for}}\:\boldsymbol{{x}}\in\right]−\pi;\boldsymbol{\pi}\right] \\ $$$$\boldsymbol{{or}}\:\boldsymbol{{sinx}}=\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$=>\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right) \\ $$$$=>\begin{cases}{\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}=\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}\pi}}\\{\frac{\pi}{\mathrm{2}}−\boldsymbol{{x}}=−\boldsymbol{{x}}−\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}}}\end{cases}\boldsymbol{{k}}\in\mathbb{Z} \\ $$$$=>\begin{cases}{\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{12}}+\boldsymbol{{k}\pi}}\\{}\end{cases} \\ $$$$\boldsymbol{{S}}_{\left.\right]\left.−\boldsymbol{\pi};\boldsymbol{\pi}\right]} =\left\{−\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{12}};\frac{\boldsymbol{\pi}}{\mathrm{12}}\right\} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:.......{le}\:{celebre}\:{cedric}\:{junior}....... \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com