急急急急!!

(a)
-1≦sin2θ≦1
0≦|sin2θ|≦1
0≦y≦1

(b)
-1≦sinθ≦1
-2≦2sinθ≦2
-7≦2sinθ-5≦-3
0≦|2sinθ-5|≦7
0≦y≦7

(c)
-1≦sinθ≦1
1≦sinθ+2≦3
1≦(sinθ+2)^2≦9
-3≦sin^2θ + 4sinθ≦5
-3≦y≦5

(d)
-1≦sinθ≦1
-2≦-2sinθ≦2
-1≦1-2sinθ≦3
1≦(1-2sinθ)^2≦9
0≦4sin^2 θ - 4sinθ≦8
-2≦-sin^2 θ + sinθ≦0
0≦1-sin^2 θ + sinθ+1≦2
0≦cos^2 θ -sinθ +1≦2
0≦y≦2

2007-01-11 21:26:02 補充：
(d)-1≦sinθ≦1-2≦-2sinθ≦2-1≦1-2sinθ≦30≦(1-2sinθ)^2≦9-1≦4sin^2 θ - 4sinθ≦8-2≦-sin^2 θ sinθ≦1/40≦1-sin^2 θ sinθ 1≦9/40≦cos^2 θ -sinθ 1≦9/40≦y≦9/4

a) -1 =< sin 2θ =<1 implies 0=< |sin 2θ| =<1 since |*| >= 0
b) -1 =< sinθ =<1
-2 =< 2sinθ =<2
-7 =< 2sinθ -5 =< -3
3 =< |2sinθ -5| =< 7

c) sin^2 θ +4sinθ =sin^2 θ +4sinθ +4 -4 = (sinθ +2)^2 -4
-1 =< sinθ =<1
1 =< sinθ +2 =<3
1 =< (sinθ +2)^2 =<9
-3 =< (sinθ +2)^2 -4 =<5
-3 =< sin^2 θ +4sinθ =<5

d) cos^2 θ -sinθ +1
=1 -sin^2 θ -sinθ +1
=2 -sin^2 θ -sinθ
=2 +(1/4) -(1/4)-sinθ -sin^2 θ
=(9/4) -(1/2 +sinθ)^2
Now,
-1/2 =<1/2 +sinθ =< 3/2
0 =< (1/2 +sinθ)^2 =< 9/4
0 >= -(1/2 +sinθ)^2 >= -9/4
9/4 >= (9/4) - (1/2 +sinθ)^2 >= 0
9/4 >= cos^2 θ -sinθ +1 >= 0

2007-01-11 21:41:56 補充：
嗚~~~點解成日唔夠快!!慢成20分鐘。

2007-01-13 23:41:03 補充：
哈! 係要選有錯的。

(a) Consider that sin 2θ is in between the value of -1 to 1 , thus |sin2 θ| is in between 0 and 1

(b) For max., the value of (2sinθ -5) must be negative. It is because sin θ is in between the value of -1 to 1. The max value of |2sinθ -5| = |2(-1) -5| = 7. On the other hand, the min. value of |2sinθ -5| = |2(1) -5| = 3.

(c) By completing the square,
y=sin^2 θ +4sinθ
=> y + 4 =sin^2 θ +4sinθ + 4
=> y + 4 = ( sinθ + 2 )^2
=> y = ( sinθ + 2 )^2 - 4
therefore the min. of value of y is ( -1 + 2 )^2 - 4 = -3
the max. value of y is ( 1 + 2 )^2 - 4 = 5

(d) y= cos^2 θ -sinθ +1
=>y = 1 - sin^2θ + sinθ +1
= -sin^2θ + sinθ + 2
= -(sinθ + 1/2)^2 + 9/4
therefore the max. of value of y is -( 0 + 1/2 )^2 +9/4 = 2
the min. value of y is -( 1 + 1/2 )^2 + 9/4 = 0

(a)
the value of sin is between -1 and 1
sin (90)=1 and sin (270)=-1
so
the range of the values of y is 0<=y<=1
(b)
Similarly, the maximum value of y is equal to |-2-5|=7
the minimum value of y is equal to |2-5|=3
the range of the values of y is 3<=y<=7
(c)
y
=sin^2 θ +4sinθ
=sin^2θ +4sinθ +4-4
=(sinθ+2)^2-4
So the maximum value of y is equal to 5
the minimum value of y is equal to -4
the range of the values of y is -4<=y<=5
(d)
y
= cos^2 θ -sinθ +1
=1-sin^2θ-sinθ+1
=-(sin^2θ+sinθ-2)
=-[(sinθ+1/2)^2-9/4]
=-(sinθ+1/2)^2+9/4
So the maximum value of y is equal to 9/4 (when sinθ=-1/2)
the minimum value of y is equal to 0
the range of the values of y is 0<=y<=9/4




2007-01-11 21:30:12 補充：
(c)think wrongthe minimum value of y is equal to -3 (when sinθ=-1)so -3&lt;=y&lt;=5