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Question Number 179269 by mnjuly1970 last updated on 27/Oct/22
what is the periodicity of thefollowing function? f(x)=cot(x)−tan(x)−2tan(2x)−4tan(4x)
$$ \\ $$$$\:\:\:{what}\:{is}\:{the}\:{periodicity}\: \\ $$$${of}\:{thefollowing}\:{function}? \\ $$$$ \\ $$$$\:{f}\left({x}\right)={cot}\left({x}\right)−{tan}\left({x}\right)−\mathrm{2}{tan}\left(\mathrm{2}{x}\right)−\mathrm{4}{tan}\left(\mathrm{4}{x}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 27/Oct/22
T=(π/8) ?
$${T}=\frac{\pi}{\mathrm{8}}\:? \\ $$
Commented by mnjuly1970 last updated on 28/Oct/22
yes sir
$$\:{yes}\:\:{sir} \\ $$
Answered by Acem last updated on 28/Oct/22
Note: Check the other method below “Answer2” determinant ((( x),(→0^− ),(→0^+ ),(π/4),(→ (π^( −) /2)),(→(π^+ /2)),(→π^− )),(( cot x),(→−∞),(→+∞),1,0,0,(−∞)),(( tan x),( 0),0,1,(+∞),(−∞),0),((g(x)_(cot x−tan x) ),(−∞),(+∞),0,(−∞),(+∞),(−∞))) ⇒ g(x) ∈ ]−∞, +∞[ for x∈ ]0, (π/2)[ : Period.= (π/2) k determinant ((( x),(→0^+ ),(π/8),(→(π^− /4)),(→(π^+ /4))),(( g(x)),(+∞),2,( 0),( 0)),(( 2tan 2x),( 0),2,(+∞),(−∞)),((M(x)_(g(x)−2tan 2x) ),(+∞),0,(−∞),(+∞))) ⇒ M(x) ∈ ]−∞, +∞[ for x∈ ]0, (π/4)[ : Period.= (π/4) k determinant ((( x),(→0^+ ),(π/(16)),(→(π^− /8)),(→(π^+ /8))),((M(x)),(+∞),4,( 0),( 0)),((4tan 4x),( 0),4,(+∞),(−∞)),(( f(x)),(+∞),0,(−∞),(+∞))) ⇒ f(x) ∈ ]−∞, +∞[ for x∈ ]0, (π/8)[ : Period.= (π/8) k ; k∈ Z f(x)= 0∣_(x= (π/(16))) f(x) →_(x→(π^( −) /8) + (π/8) k) ^(x→0^+ + (π/8) k) ±∞
$${Note}:\:{Check}\:{the}\:{other}\:{method}\:{below}\:“\boldsymbol{{Answer}}\mathrm{2}'' \\ $$$$ \\ $$$$\:\begin{array}{|c|c|c|c|}{\:\:\:\:\:{x}}&\hline{\rightarrow\mathrm{0}^{−} }&\hline{\rightarrow\mathrm{0}^{+} }&\hline{\frac{\pi}{\mathrm{4}}}&\hline{\rightarrow\:\frac{\pi^{\:−} }{\mathrm{2}}}&\hline{\rightarrow\frac{\pi^{+} }{\mathrm{2}}}&\hline{\rightarrow\pi^{−} }\\{\:\:\:\:\mathrm{cot}\:{x}}&\hline{\rightarrow−\infty}&\hline{\rightarrow+\infty}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{−\infty}\\{\:\:\:\:\mathrm{tan}\:{x}}&\hline{\:\:\:\:\:\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}&\hline{+\infty}&\hline{−\infty}&\hline{\mathrm{0}}\\{{g}\left({x}\underset{\mathrm{cot}\:{x}−\mathrm{tan}\:{x}} {\right)}}&\hline{−\infty}&\hline{+\infty}&\hline{\mathrm{0}}&\hline{−\infty}&\hline{+\infty}&\hline{−\infty}\\\hline\end{array} \\ $$$$\left.\:\:\Rightarrow\:{g}\left({x}\right)\:\in\:\right]−\infty,\:+\infty\left[\:{for}\:{x}\in\:\right]\mathrm{0},\:\frac{\pi}{\mathrm{2}}\left[\::\:{Period}.=\:\frac{\pi}{\mathrm{2}}\:{k}\right. \\ $$$$\: \\ $$$$\:\begin{array}{|c|c|c|c|}{\:\:\:\:\:\:\:\:\:{x}}&\hline{\rightarrow\mathrm{0}^{+} }&\hline{\frac{\pi}{\mathrm{8}}}&\hline{\rightarrow\frac{\pi^{−} }{\mathrm{4}}}&\hline{\rightarrow\frac{\pi^{+} }{\mathrm{4}}}\\{\:\:\:\:\:\:{g}\left({x}\right)}&\hline{+\infty}&\hline{\mathrm{2}}&\hline{\:\:\:\:\mathrm{0}}&\hline{\:\:\:\mathrm{0}}\\{\:\:\:\mathrm{2tan}\:\mathrm{2}{x}}&\hline{\:\:\:\:\mathrm{0}}&\hline{\mathrm{2}}&\hline{+\infty}&\hline{−\infty}\\{{M}\left({x}\underset{{g}\left({x}\right)−\mathrm{2tan}\:\mathrm{2}{x}} {\right)}}&\hline{+\infty}&\hline{\mathrm{0}}&\hline{−\infty}&\hline{+\infty}\\\hline\end{array} \\ $$$$\left.\:\:\Rightarrow\:{M}\left({x}\right)\:\in\:\right]−\infty,\:+\infty\left[\:{for}\:{x}\in\:\right]\mathrm{0},\:\frac{\pi}{\mathrm{4}}\left[\::\:{Period}.=\:\frac{\pi}{\mathrm{4}}\:{k}\right. \\ $$$$ \\ $$$$\:\begin{array}{|c|c|c|c|}{\:\:\:\:\:{x}}&\hline{\rightarrow\mathrm{0}^{+} }&\hline{\frac{\pi}{\mathrm{16}}}&\hline{\rightarrow\frac{\pi^{−} }{\mathrm{8}}}&\hline{\rightarrow\frac{\pi^{+} }{\mathrm{8}}}\\{{M}\left({x}\right)}&\hline{+\infty}&\hline{\mathrm{4}}&\hline{\:\:\:\:\mathrm{0}}&\hline{\:\:\:\:\mathrm{0}}\\{\mathrm{4tan}\:\mathrm{4}{x}}&\hline{\:\:\:\:\mathrm{0}}&\hline{\mathrm{4}}&\hline{+\infty}&\hline{−\infty}\\{\:\:\:{f}\left({x}\right)}&\hline{+\infty}&\hline{\mathrm{0}}&\hline{−\infty}&\hline{+\infty}\\\hline\end{array} \\ $$$$\left.\:\:\Rightarrow\:{f}\left({x}\right)\:\in\:\right]−\infty,\:+\infty\left[\:{for}\:{x}\in\:\right]\mathrm{0},\:\frac{\pi}{\mathrm{8}}\left[\::\:{Period}.=\:\frac{\pi}{\mathrm{8}}\:{k}\right. \\ $$$$\:;\:{k}\in\:\mathbb{Z} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\:\mathrm{0}\mid_{{x}=\:\frac{\pi}{\mathrm{16}}} \:\:\:\:\:\:\:\:\:{f}\left({x}\right)\:\underset{{x}\rightarrow\frac{\pi^{\:−} }{\mathrm{8}}\:+\:\frac{\pi}{\mathrm{8}}\:{k}} {\overset{{x}\rightarrow\mathrm{0}^{+} +\:\frac{\pi}{\mathrm{8}}\:{k}} {\rightarrow}}\:\pm\infty\:\: \\ $$
Commented by Acem last updated on 27/Oct/22
why is the pic i uploaded not shwoing How to upload pictures on comments ): ?
$${why}\:{is}\:{the}\:{pic}\:{i}\:{uploaded}\:{not}\:{shwoing} \\ $$$$ \\ $$$$\left.{How}\:{to}\:{upload}\:{pictures}\:{on}\:{comments}\:\right):\:? \\ $$
Commented by mr W last updated on 27/Oct/22
what is the proof for ((πk)/8), ((πk)/4) etc.? is it obvious?
$${what}\:{is}\:{the}\:{proof}\:{for}\:\frac{\pi{k}}{\mathrm{8}},\:\frac{\pi{k}}{\mathrm{4}}\:{etc}.? \\ $$$${is}\:{it}\:{obvious}? \\ $$
Commented by Acem last updated on 28/Oct/22
Ever, that is why i was asking about uploading pictures, I draw graphs on a paper explaining the sum of functions... after dinner will organize a table on here
$${Ever},\:{that}\:{is}\:{why}\:{i}\:{was}\:{asking}\:{about}\:{uploading} \\ $$$$\:{pictures},\:{I}\:{draw}\:{graphs}\:{on}\:{a}\:{paper}\:{explaining} \\ $$$$\:{the}\:{sum}\:{of}\:{functions}…\:{after}\:{dinner}\:{will} \\ $$$$\:{organize}\:{a}\:{table}\:{on}\:{here} \\ $$
Commented by Acem last updated on 28/Oct/22
I noted tables above, although i prefer graphs
$${I}\:{noted}\:{tables}\:{above},\:{although}\:{i}\:{prefer}\:{graphs} \\ $$
Commented by Acem last updated on 28/Oct/22
Also there′s other method, it′s bellow
$${Also}\:{there}'{s}\:{other}\:{method},\:{it}'{s}\:{bellow} \\ $$
Answered by Acem last updated on 28/Oct/22
With Algebra: f(x)= cot x− tan x −2 tan 2x −4 tan 4x = 2 cot 2x −2 tan 2x −4 tan 4x = 4 cot 4x −4 tan 4x f(x) = 8 cot 8x ⇒ The periodicity= (π/8) ; 8x= π+πk i.e. x= (𝛑/8) + (π/8) k ; k∈ Z
$${With}\:{Algebra}: \\ $$$$\:\:\:{f}\left({x}\right)=\:\mathrm{cot}\:{x}−\:\mathrm{tan}\:{x}\:−\mathrm{2}\:\mathrm{tan}\:\mathrm{2}{x}\:−\mathrm{4}\:\mathrm{tan}\:\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cot}\:\mathrm{2}{x}\:−\mathrm{2}\:\mathrm{tan}\:\mathrm{2}{x}\:−\mathrm{4}\:\mathrm{tan}\:\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}\:\mathrm{cot}\:\mathrm{4}{x}\:−\mathrm{4}\:\mathrm{tan}\:\mathrm{4}{x} \\ $$$$\:{f}\left({x}\right)\:\:=\:\mathrm{8}\:\mathrm{cot}\:\mathrm{8}{x} \\ $$$$\:\Rightarrow\:{The}\:{periodicity}=\:\frac{\pi}{\mathrm{8}}\: \\ $$$$\:;\:\mathrm{8}{x}=\:\pi+\pi{k}\:\:\:\:{i}.{e}.\:\:{x}=\:\frac{\boldsymbol{\pi}}{\mathrm{8}}\:+\:\frac{\pi}{\mathrm{8}}\:{k}\:;\:{k}\in\:\mathbb{Z} \\ $$
Commented by mr W last updated on 28/Oct/22
i think it causes more confusion when you write 8x=π+πk ⇒x=(π/8)+((πk)/8). but when you think this is more clear, you can do.
$${i}\:{think}\:{it}\:{causes}\:{more}\:{confusion}\:{when} \\ $$$${you}\:{write}\:\mathrm{8}{x}=\pi+\pi{k}\:\Rightarrow{x}=\frac{\pi}{\mathrm{8}}+\frac{\pi{k}}{\mathrm{8}}. \\ $$$${but}\:{when}\:{you}\:{think}\:{this}\:{is}\:{more}\:{clear}, \\ $$$${you}\:{can}\:{do}. \\ $$
Commented by Acem last updated on 28/Oct/22
cot breaks tan′s head
$$\mathrm{cot}\:{breaks}\:\mathrm{tan}'{s}\:{head} \\ $$
Commented by mr W last updated on 28/Oct/22
nice! this is the method i used. cot x−tan x=((cos x)/(sin x))−((sin x)/(cos x)) =((cos^2 x−sin^2 x)/(sin x cos x))=((2 cos 2x)/(sin 2x)) cot x−tan x−2 tan 2x =((2 cos 2x)/(sin 2x))−((2 sin 2x)/(cos 2x))=((4 cos 4x)/(sin 4x)) cot x−tan x−2 tan 2x−4 tan 4x =((4 cos 4x)/(sin 4x))−((4 sin 4x)/(cos 4x))=((8 cos 8x)/(sin 8x)) =8 cot 8x=f(x) f(x+T)=f(x) 8(x+T)=8x+kπ ⇒T=((kπ)/8) ⇒period is (π/8).
$${nice}! \\ $$$${this}\:{is}\:{the}\:{method}\:{i}\:{used}. \\ $$$$\mathrm{cot}\:{x}−\mathrm{tan}\:{x}=\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$\:\:\:=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$\mathrm{cot}\:{x}−\mathrm{tan}\:{x}−\mathrm{2}\:\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\:\:\:=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}−\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}=\frac{\mathrm{4}\:\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\mathrm{cot}\:{x}−\mathrm{tan}\:{x}−\mathrm{2}\:\mathrm{tan}\:\mathrm{2}{x}−\mathrm{4}\:\mathrm{tan}\:\mathrm{4}{x} \\ $$$$\:\:\:=\frac{\mathrm{4}\:\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{sin}\:\mathrm{4}{x}}−\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{cos}\:\mathrm{4}{x}}=\frac{\mathrm{8}\:\mathrm{cos}\:\mathrm{8}{x}}{\mathrm{sin}\:\mathrm{8}{x}} \\ $$$$\:\:\:=\mathrm{8}\:\mathrm{cot}\:\mathrm{8}{x}={f}\left({x}\right) \\ $$$${f}\left({x}+{T}\right)={f}\left({x}\right) \\ $$$$\mathrm{8}\left({x}+{T}\right)=\mathrm{8}{x}+{k}\pi \\ $$$$\Rightarrow{T}=\frac{{k}\pi}{\mathrm{8}}\:\Rightarrow{period}\:{is}\:\frac{\pi}{\mathrm{8}}. \\ $$
Commented by mnjuly1970 last updated on 28/Oct/22
very nice as always sir W
$$\:\:{very}\:{nice}\:{as}\:{always}\:{sir}\:{W} \\ $$
Commented by mnjuly1970 last updated on 28/Oct/22
thank you so much sir
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$
Commented by Tawa11 last updated on 28/Oct/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Acem last updated on 28/Oct/22
Yes friend, it′s more reliable.
$${Yes}\:{friend},\:{it}'{s}\:{more}\:{reliable}. \\ $$
Commented by Acem last updated on 28/Oct/22
friend, this 8(x+T)=8x+kπ causes a bit of confusion, it′s better to write like this: f(x)= 8 cot (8x) 8x= π+πk ⇒ x= (𝛑/8) + (π/8)k
$${friend},\:{this}\:\mathrm{8}\left({x}+{T}\right)=\mathrm{8}{x}+{k}\pi\:{causes}\:{a}\:{bit} \\ $$$$\:{of}\:{confusion},\:{it}'{s}\:{better}\:{to}\:{write}\:{like}\:{this}: \\ $$$$\:{f}\left({x}\right)=\:\mathrm{8}\:\mathrm{cot}\:\left(\mathrm{8}\boldsymbol{{x}}\right) \\ $$$$\:\mathrm{8}{x}=\:\pi+\pi{k}\:\Rightarrow\:{x}=\:\frac{\boldsymbol{\pi}}{\mathrm{8}}\:+\:\frac{\pi}{\mathrm{8}}{k} \\ $$$$ \\ $$
Commented by Acem last updated on 28/Oct/22
I have no problem with both, both are nice
$${I}\:{have}\:{no}\:{problem}\:{with}\:{both},\:{both}\:{are}\:{nice} \\ $$

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