The maximum of function y = 3sin (x / 3) + 4cos (x / 3)

The maximum of function y = 3sin (x / 3) + 4cos (x / 3)


y=3sin(x/3)+4cos(x/3)
=5[(3/5)sin(x/3)+(4/5)cos(x/3)]
=5sin (x / 3 + α) (where cos α = 3 / 5, sin α = 4 / 5)
∈[-5,5]



It is known that the maximum value of the function y = 3sin α + 4cos α + A is 7, then a=


asinx+bcosx=√(a²+b8)sin(x+y)
Where tany = B / A
Here a = 3, B = 4
So y = 5sin (α + β) + a max = 5 + a = 7
a=2

Elden Ring Premiere

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