S.4~ TRIGO. QUESTION!! HELP~

Given
that 0˚<=x<=360˚

(a) Find the maximum value of (3-2sinx) and its corresponding measure of x.

－1<=Sinx<=1

－1<=－sinx<=1

－2<=-2Sinx<=2

3－2<=3-2Sinx<=3+2

1<=3-2Sinx<=5

when mmax 5=3+2Sinx

Sinx=1

x=90˚

So max=5，x=90˚



(b) Find the minimum value of (3+2Sinx) and its corresponding measure of x.

Sol

－1<=Sinx<=1

－2<=2Sinx<=2

3－2<=3+2Sinx<=3+2

1<=3+2Sinx<=5

when min 1=3+2Sinx

Sinx=－1

x=270˚

So min=1，x=270˚



(c) Find the maximum value of (3－2Sinx)/(3+2Sinx)

Sol

A=(3－2Sinx)/(3+2Sinx)

3A+2ASinx=3－2Sinx

(2A+2)Sinx=3－3A

Sinx=(3－3A)/(2A+2)

｜(3－3A)/(2A+2)｜<=1

｜3－3A｜<=｜2A+2｜

9A^2－18A+9<=4A^2+8A+4

5A^2－26A+5<=0

(A－5)(5A－1)<=0

(A－5)(A－1/5)<=0

1/5<=A<=5

max=5







2010-07-13 15:56:39 補充：
(c) Find the maximum value of (3-2sinx)/(3+2sinx)
when sinx=1
(3-2sinx)/(3+2sinx)=1/5
So min<>1
.

a)Find the maximum value of (3-2sinx) and its corresponding measure of x.
0˚≤x≤360˚
-1≤sinx≤1
2≥-2sinx≥-2
5≥3-2sinx≥-3
so the maximum value of (3-2sinx) is 5

(3-2sinx)=5
(-2sinx)=2
sinx=-1
x=180˚+sin-1-1 or x=360˚ - sin-1-1
x=270˚


(b) Find the minimum value of (3+2sinx) and its corresponding measure of x.
0˚≤x≤360˚
-1≤sinx≤1
-2≤2sinx≤2
1≤3+2sinx≤5
so the minimum value of (3+2sinx) is 1

(3+2sinx)=1
(2sinx)=-2
sinx=-1
x=180˚+sin-1-1 or x=360˚ - sin-1-1
x=270˚

(c) Find the maximum value of 3-2sinx/3+2sinx.
1≤3+2sinx≤5
1≥1/(3+2sinx) ≥1/5 FROM b) know 5≥3-2sinx≥-3
1(5) ≥(3-2sinx)/(3+2sinx) ≥5(1/5)
5≥(3-2sinx)/(3+2sinx) ≥1
so the maximum value of 3-2sinx/3+2sinx is5