What is the frequency of #f(theta)= sin 18 t - cos 9 t #?

1 Answer
Jul 16, 2018

The frequency is #f=9/(2pi)Hz#

Explanation:

First determine the period #T#

The period #T# of a periodic function #f(x)# is defined by

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

Here,

#f(t)=sin(18t)-cos(9t)#............................#(1)#

Therefore,

#f(t+T)=sin(18(t+T))-cos(9(t+T))#

#=sin(18t+18T)-cos(9t+9T)#

#=sin18tcos18T+cos18Tsin18t-cos9tcos9T+sin9tsin9T#

Comparing #f(t)# and #f(t+T)#

#{(cos18T=1),(sin18T=0),(cos9T=1),(sin9T=0):}#

#<=>#, #{(18T=2pi),(9T=2pi):}#

#=>#, #T_1=pi/9# and #T_2=2/9pi#

The #LCM# of #T_1# and #T_2# is #T=2/9pi#

Therefore,

The frequency is

#f=1/T=9/(2pi)Hz#

graph{sin(18x)-cos(9x) [-2.32, 4.608, -1.762, 1.703]}