Let, #f(x)=13cosx+3sqrt3sinx-4#.
#:. f(x)=14(13/14cosx+(3sqrt3)/14sinx)-4............(star)#.
Note that, #(13/14)^2+((3sqrt3)/14)^2=169/14^2+27/14^2=196/14^2=1#.
This means that, the point #(13/14,(3sqrt3)/14)# lies on the
unit circle : #x^2+y^2=1#.
Consequently, #EE# a unique #theta in [0,2pi)#, such that,
#costheta=13/14 and sintheta=(3sqrt3)/14............(star^star)#.
Utilising this fact in #(star)#, we have,
#f(x)=14{costhetacosx+sinthetasinx}-4, or, #
#f(x)=14cos(x-theta)-4#.
But, #AA x in RR, -1 le cos(x-theta) le 1#.
Multiplying by #14 gt 0, -14 le 14cos(x-theta) le 14, x in RR#.
Adding #-4, -18 le 14cos(x-theta)-4 le 10, x in RR#.
# rArr AA x in RR, -18 le f(x) le 10#.
# rArr f_min=-18 and f_max=10#.
