What is the period of #f(t)=sin( t /13 )+ cos( (13t)/24 ) #?

2 Answers
Narad T.
Jun 11, 2018

The period is #=4056pi#

Explanation:

The period #T# of a periodic functon is such that

#f(t)=f(t+T)#

Here,

#f(t)=sin(1/13t)+cos(13/24t)#

Therefore,

#f(t+T)=sin(1/13(t+T))+cos(13/24(t+T))#

#=sin(1/13t+1/13T)+cos(13/24t+13/24T)#

#=sin(1/13t)cos(1/13T)+cos(1/13t)sin(1/13T)+cos(13/24t)cos(13/24T)-sin(13/24t)sin(13/24T)#

As,

#f(t)=f(t+T)#

#{(cos(1/13T)=1),(sin(1/13T)=0), (cos(13/24T)=1),(sin(13/24T)=0):}#

#<=>#, #{(1/13T=2pi),(13/24T=2pi):}#

#<=>#, #{(T=26pi=338pi),(T=48/13pi=48pi):}#

#<=>#, #T=4056pi#

Nghi N.
Jun 11, 2018

#624pi#

Explanation:

Period of #sin (t/13)# --> #13(2pi) = 26pi#
Period of #cos ((13t)/24)# --> #((24)(2pi))/13 = (48pi)/13#
Period of f(t) --> least common multiple of #26pi# and #(48pi)/13#

#26pi# .... x (24).............--> .#624pi#
#(48pi)/13# .....x (13)(13)...--> #624pi#...-->
Period of f(t) --> #624pi#

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