how ?
cos(3x+(pi/3))={[1/2]*cos3x-[sqr(3)/2]*sin3x}
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
so
cos(3x + (π/3))
= cos(3x)cos(π/3) - sin(3x)sin(π/3)
and cos(π/3) = 1/2 while sin(π/3) = (√3)/2
= cos(3x)/2 - (sin(3x)*√3)/2
as per your given answer
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cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
so
cos(3x + (π/3))
= cos(3x)cos(π/3) - sin(3x)sin(π/3)
and cos(π/3) = 1/2 while sin(π/3) = (√3)/2
= cos(3x)/2 - (sin(3x)*√3)/2
as per your given answer