Question

find the maximum and minimum value of y=3sin x - 4cos x, where x belongs to [0, π/2]​

Answer 1

Answer:

Maximum Value =5

Minimum value=-5

Step-by-step explanation:

Given y=3sinx-4cosx

Let y=asinx+bcosx+c

a=3

b= -4

c=0

Maximum value=c+sqrt (a^2+b^2)

=0+sqrt (3^2+(-4)^2)

=0+sqrt (25)

Maximum Value=5

Minimum Value=c-sqrt (a^2+b^2)

=0-5

=-5

Answer 2

answer

maximus value =5 minimum value = 5

step by step explanation

given y = 3 sin x - 4 cosx

let y = asin x + becox+ c

a= 3 b= 4 c= 0

maximus value =c+ sqrt (a^2+ b^2)

= 0 + sqrt (3^+(-4)^2)

= 0+ sqrt (25)

maximus value 5

minimum value = c- sqrt (a^2+b^2)

= 0-5

= 5

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