Find max and min values of y=sinx sin2x

Answer 1

y = sinx sin2x
y' = cosx sin2x + 2sinx cos2x
= 2sinx cos^2x + 2sinx - 4sin^3x
= 2sinx - 2sin^3x + 2sinx - 4sin^3x
= 4sinx - 6sin^3x
= 2sinx (2 - 3sin^2x)

so, max/min occurs when sinx = √(2/3)
When six = √(2/3), cosx = √(1/3)
so, sin2x = √(8/9)

max/min of y is thus ±√(16/27) = 4√3/9 = .7698