What is the period of #f(t)=sin( t /6 )+ cos( (7t)/24 ) #?

2 Answers
Narad T.
May 12, 2018

The period is #=12pi#

Explanation:

A periodic function #f(x)# is such that

#f(x)=f(x+T)#

where, #T# is the period

Here,

#f(t)=sin(t/6)+cos(7/24t)#

#f(t+T)=sin(1/6(t+T))+cos(7/24(t+T))#

#=sin(1/6t+1/6T)+cos(7/24t+7/24T)#

Therefore,

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

#{(sin(t/6)=sin(1/6t+1/6T)),(cos(7/24t)=cos(7/24t+7/24T)):}#

#<=>#, #{(sin(t/6)=sin(t/6)cos(1/6T)+cos(t/6)sin(1/6T)),(cos(7/24t)=cos(7/24t)cos(7/24T)-sin(7/24t)sin(7/24T)):}#

#<=>#, #{(cos(1/6T)=1),(sin(1/6T)=0),(cos(7/24T)=1),(sin(7/24T)=0):}#

#<=>#, #{(1/6T=2pi),(7/24T=2pi):}#

#<=>#, #{(T=12pi),(T=48/7pi=48pi):}#

The LCM of #12pi# and #48pi# is #=12pi#

The period is #=12pi#

graph{sin(x/6)+cos(7x/24) [-8.3, 56.63, -15.6, 16.88]}

Nghi N
May 13, 2018

#48pi#

Explanation:

Period of #sin (t/6) --> 6(2pi) = 12pi#
Period of #cos ((7t)/24) --> (24(2pi))/7 = (48pi)/7#
Period of f(t) is the least common multiple of #12pi and (48pi)/7#.
#12pi# .....x (4) ........ --> #48pi#
#(48pi)/7# ...x (7) .... --> #48 pi#

Period of f(t) --> #48pi#

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