F.4 M2 TRIGONOMETRY

A)3sin2x - √3 cos2x= 2√3 ( (√3/2) sin2x - (1/2) cos2x )= 2√3 ( cos π/6 sin2x - sin π/6 cos2x )= 2√3 sin (2x - π/6) = 2√3 cos (π/2 - (2x - π/6)) = 2√3 cos (2π/3 - 2x)= 2√3 cos (2x - 2π/3)
B)y =1 / (6sinx cosx + 2√3 sin²x - 4√3)y = 1 / (3 sin2x + √3 (2sin²x - 4) )y = 1 / (3 sin2x + √3 (1 - cos2x - 4) )y = 1 / (3 sin2x - √3 cos2x - 3√3)By the result of A) ,y = 1 / (2√3 cos (2x - 2π/3) - 3√3)Thus , y (max.) = 1 / ( - 2√3 - 3√3) = - √3 / 5y (min.) = 1 / ( 2√3 - 3√3) = - √3 / 3


2011-07-26 16:57:10 補充：
Sorry!

y (max.) should be = 1 / ( - 2√3 - 3√3) = - √3 / 15

2011-07-26 17:21:06 補充：
cos (2x - 2π/3) 最大 = 1 , 最小 = - 1

所以分母 (2√3 cos (2x - 2π/3) - 3√3) 最大 = 2√3 - 3√3 = - √3

最小 = - 2√3 - 3√3 = - 5√3

對負分母而言 , 1 / - 5√3 最大 ,

1 / - √3 最小。

A)3sin2x -√3 cos2x
＝2√3( 3/[2√3]sin2x -√3/[2/√3]cos2x)
=2√3(sin(60)sin2x-cos(60)cos2x)
=2√3(sin(180-60)sin2x + cos(180-60)(cos2x))
=2/√3(sin(2π/3)sin2x+cos(2π/3)cos2x)
=2/√3(cos(2x-2π/3) (by cos(A-B)=cosAcosB+sinAsinB)

B) first consider (6sinx cosx + 2√3 sin^2 x -4√3)
(6sinx cosx + 2√3 sin^2 x -4√3)

=3sin2x+2√3 sin^2 x -4√3
=3sin2x+2√3((1-cos2x)/2) -4√3
=3sin2x-√3cos2x+√3-4√3

by the result of part A, we have
=2√3cos(2x-2π/3)-3√3

it attains its highest value when cos(2x-2π/3)=1,
the maximum value of 2√3cos(2x-2π/3)-3√3 = -√3

and it attains its lowest when cos(2x-2π/3)=-1
the minium value of 2√3cos(2x-2π/3)-3√3 = -5√3

therefore the extremum of y =-1/√3 and -1/(5√3)

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