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Question Number 183563 by CrispyXYZ last updated on 27/Dec/22

repost an unsolved question Q182552 Find the period of the following: a• sin 4x sin 3x b• sin πx+ cos x c• ((2 sin^2 3x− 3 tan 4x+ 4 cot 6x)/(∣cosec 8x∣− sec^3 10x+ (√(cot 12x))))

$$\mathrm{repost}\:\mathrm{an}\:\mathrm{unsolved}\:\mathrm{question}\:\mathrm{Q182552} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}: \\ $$$$\:{a}\bullet\:\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\:{b}\bullet\:\mathrm{sin}\:\pi{x}+\:\mathrm{cos}\:{x} \\ $$$$\:{c}\bullet\:\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}{x}−\:\mathrm{3}\:\mathrm{tan}\:\mathrm{4}{x}+\:\mathrm{4}\:\mathrm{cot}\:\mathrm{6}{x}}{\mid\mathrm{cosec}\:\mathrm{8}{x}\mid−\:\mathrm{sec}^{\mathrm{3}} \:\mathrm{10}{x}+\:\sqrt{\mathrm{cot}\:\mathrm{12}{x}}} \\ $$

Commented by Acem last updated on 27/Dec/22

Thanks friend for reminding us this question, hope from all sharing their methodes.

$${Thanks}\:{friend}\:{for}\:{reminding}\:{us}\:{this}\:{question}, \\ $$$$\:{hope}\:{from}\:{all}\:{sharing}\:{their}\:{methodes}. \\ $$

Answered by mr W last updated on 27/Dec/22

(a) f(x)=sin 4x sin 3x 4T=2nπ ⇒T=((3nπ)/6) 3T=2mπ ⇒T=((4mπ)/6) T_(min) =2π ✓ (b) f(x)=sin πx+cos x πT=2nπ ⇒T=2n=even integer ⇒T=2mπ=even integer ×π ⇒no periodic function! (c) 2 sin^2 3x=1−cos 6x 6T=2aπ ⇒T=((aπ×20)/(60)) 8T=2bπ ⇒T=((bπ×15)/(60)) 10T=2cπ ⇒T=((cπ×12)/(60)) 4T=dπ ⇒T=((3dπ)/(12)) 6T=eπ ⇒T=((2eπ)/(12)) 12T=fπ ⇒T=((fπ)/(12)) ⇒T_(min) =π ✓

$$\left({a}\right) \\ $$$${f}\left({x}\right)=\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{4}{T}=\mathrm{2}{n}\pi\:\Rightarrow{T}=\frac{\mathrm{3}{n}\pi}{\mathrm{6}} \\ $$$$\mathrm{3}{T}=\mathrm{2}{m}\pi\:\Rightarrow{T}=\frac{\mathrm{4}{m}\pi}{\mathrm{6}} \\ $$$${T}_{{min}} =\mathrm{2}\pi\:\checkmark \\ $$$$\left({b}\right) \\ $$$${f}\left({x}\right)=\mathrm{sin}\:\pi{x}+\mathrm{cos}\:{x} \\ $$$$\pi{T}=\mathrm{2}{n}\pi\:\Rightarrow{T}=\mathrm{2}{n}={even}\:{integer} \\ $$$$\Rightarrow{T}=\mathrm{2}{m}\pi={even}\:{integer}\:×\pi \\ $$$$\Rightarrow{no}\:{periodic}\:{function}! \\ $$$$\left({c}\right) \\ $$$$\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}{x}=\mathrm{1}−\mathrm{cos}\:\mathrm{6}{x} \\ $$$$\mathrm{6}{T}=\mathrm{2}{a}\pi\:\Rightarrow{T}=\frac{{a}\pi×\mathrm{20}}{\mathrm{60}} \\ $$$$\mathrm{8}{T}=\mathrm{2}{b}\pi\:\Rightarrow{T}=\frac{{b}\pi×\mathrm{15}}{\mathrm{60}} \\ $$$$\mathrm{10}{T}=\mathrm{2}{c}\pi\:\Rightarrow{T}=\frac{{c}\pi×\mathrm{12}}{\mathrm{60}} \\ $$$$\mathrm{4}{T}={d}\pi\:\Rightarrow{T}=\frac{\mathrm{3}{d}\pi}{\mathrm{12}} \\ $$$$\mathrm{6}{T}={e}\pi\:\Rightarrow{T}=\frac{\mathrm{2}{e}\pi}{\mathrm{12}} \\ $$$$\mathrm{12}{T}={f}\pi\:\Rightarrow{T}=\frac{{f}\pi}{\mathrm{12}} \\ $$$$\Rightarrow{T}_{{min}} =\pi\:\checkmark \\ $$

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