#f(t) = sin 5t - cos 35 t #. Let
#p_1# = period of #sin 5t = (2pi)/5 and#
#p_2# = period of #- cos 35t = (2pi)/35#
Now,
the period ( least possible ) P of #f ( t )# has to be satisfy
#P = p_1L + p_2M#
#= 2/5 L pi = 2/35M# such tjat
#f ( t + P ) = f ( t )#
As 5 is a factor of 35, their LCM = 35 and
#35 P =14Lpi = 2Mpi rArr L =1, M = 7 and P= 14/35pi = 2/5pi#
See that #f ( t + 2/5pi) = sin (5t+ 2pi) - cos ( 35 t + 14 pi )#
#= sin4t -cos 35t = f (t )# and that
#f ( t + P/2 ) = sin ( 5t + pi ) - cos ( 35t + 7pi )#
#= - sin 5t + cos 35t #
#ne f ( t )#
See graph.
graph{(y- sin (5x) + cos (35x))(x-pi/5 +.0001y)(x+pi/5 +0.0001y)=0[-1.6 1.6 -2 2]}
Observe the lines #x = +-pi/5 = +-0.63#, nearly, to mark the period.
For better visual effect, the graph is not on uniform scale.
