If the maximum value of the function y = a SiNx + 3cosx + 1 is 6, then a=
Y = √ (a ^ 2 + 3 ^ 2) sin (x + T) + 1, where Tan = 3 / A
Maximum = √ (a ^ 2 + 3 ^ 2) + 1 = 6
That is, a ^ 2 + 3 ^ 2 = 25
We get a = 4 or - 4
Find the maximum and minimum of the function y = 3cosx SiNx
The function y = 3cosx SiNx = 2 (32cosx-12sinx) = 2cos (x + π 6) ∈ [- 2, 2]. The maximum and minimum values of the function y = 3cosx SiNx are: 2, - 2
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