If y = 1 + 2sinx + 3 cos^2x then find minimum and maximum value of y?

If y = 1 + 2sinx + 3 cos^2x then find minimum and maximum value of y

2 Answers

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  • 7 years ago
    Favorite Answer

    y = 1 + 2sin(x) + 3 cos^2(x)

    y = 4 + 2sin(x) - 3sin^2(x)

    y = 4 - 3[ sin^2(x) - (2/3)sin(x) + (1/9)] + (1/3)

    y = (13/3) - 3[ sin(x) - (1/3)]^2

    Max value for y is 13/3

    Min value for y is for max absolute value of sin(x) - (1/3).

    Max absolute value for sin(x) - (1/3) is 4/3

    then min value for y = (13/3) - 3 * 4/3 = 1/3

    Edit:

    y = (13/3) - 3[ sin(x) - (1/3)]^2 has max value when [ sin(x) - (1/3)]^2 is zero

    y = (13/3) - 3[ sin(x) - (1/3)]^2 has min value when [ sin(x) - (1/3)]^2 has max value

    Because of ^2 expression sin(x) - (1/3) has a ABSOLUTE value -4/3 when sin(x) = -1 then

    min value is y = (13/3) - 3[ -4/3]^2 = 13/3 - 16/3 = -1

    You may use first derivative of y = 1 + 2sinx + 3 cos^2x but more laborious

  • essman
    Lv 4
    3 years ago

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