#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.