What is the frequency of #f(theta)= sin 2 t - cos 5 t #?

Question

3355 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-28T04:56:08

#2pi#

Period of sin 2t --> #(2pi)/2 = pi#
Period of cos 5t -->#(2pi)/5#
Period of f(t) --> least common multiple of #pi and (2pi)/5.#

pi .............x 2 ... --> 2pi
(2pi)/5 ....x 5 ......--> 2pi

Period of f(t) is #(2pi)#

Answer 2 · 2018-04-28T05:48:31

The frequency is #=1/(2pi)#

The frequency is #f=1/T#

The period is #=T#

A function #f(theta)# is T-periodic iif

#f(theta)=(theta+T)#

Therefore,

#sin(2t)-cos(5t)=sin2(t+T)-cos5(t+T)#

Therefore,

#{(sin (2t)=sin2(t+T)),(cos(5t)=cos5(t+T)):}#

#<=>#, #{(sin2t=sin(2t+2T)),(cos5t=cos(5t+5T)):}#

#<=>#, #{(sin2t=sin2tcos2T+cos2tsin2T),(cos5t=cos5tcos5T-sin5tsin5T):}#

#<=>#, #{(cos2T=1),(cos5T=1):}#

#<=>#, #{(2T=2pi=4pi),(5T=2pi=4pi=6pi=8pi=10pi):}#

#<=>#, #{(T=4/2pi=2pi),(T=10/5pi=2pi):}#

The period is #=2pi#

The frequency is

#f=1/(2pi)#

graph{sin(2x)-cos(5x) [-3.75, 18.75, -7.045, 4.205]}