If 0 x, find the maximum and minimum values of the function y sin2x sinx cosx

Since 0 x, -pi 4 x - pi 4 3 * pi 4

Then -(root number 2) 2 sinx(x-pi 4) (root number 2) 2y=sin2x + sinx-cosx

1-(sinx-cosx) 2+(sinx-cosx)-[sinx-cosx)-1 2] 2+5 4-[sin(x-pi 4)-1 2] 2+5 4 so when sinx(x-pi 4) = - (root number 2) 2, y minimum = -1, when sinx(x-pi 4) 1 2, y maximum = 5 4

y=sin2x+sinx-cosx

1-(sinx-cosx) 2+(sinx-cosx)5 4-(sinx-cosx-1 2) 25 4-[sin(x-pi 4)-1 2] 20 x ==> -1 2 sin(x-pi 4) 2 times the root number 2

1 4 5 4 - [root number 2 times sin(x-pi 4)-1 2] 2 5 4

i.e. 1 4 y 5 4

Therefore, the maximum value of the function y=sin2x+sinx-cosx is 5 4, and the minimum value is 1 4.

Let t=sinx+cosx=

2sin（x+π

4) [2,2], then 2sinxcosx=t2-1, then y=t2+t+1=(t+1

4,t [-2,2], the maximum value of which is 3+

2, the minimum value is 34

When x [0,

2], then t [1,2], where the maximum value of y is 3+

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y=sinx+cosx+sinxcosx

Let sinx-cosx=t,(1)

By the coangular trigonometric relationship sinxcosx=[(sinx+cosx) 2-(sinx 2+cosx 2)]2

Substituting equation (1) gives sinxcosx=(1-t 2) 2, so y=t+(1-t 2) 2

y=-1 2(t-1) 2+3 4 and because x [0, ], sinx-cosx= 2sin(x- 4) [1, 2].

So y is not monotonic at t [-1, 2].

When t = -1, y obtains the minimum value = -1

When t = 1, y gets the maximum value = 3 4

value range [-1,3 4 ].

The solution consists of y=sinx+cosx+sin2x

sinx+cosx+2sinxcosx

Let t=sinx+cosx=

Then t=sinx+cosx= 2(2 2sinx+ 2 2cosx)= 2sin(x+4).

Know-2 t 2

and by t 2 = (sinx + cosx) 2 = 1 + 2 sinxcosx, i.e. sinxcosx = (t 2-1) 2

So the original function becomes.

y=t-(t^2-1)

t^2+t+1

(t-1/2)^2+5/4

Therefore, when t=1 2, y has a maximum value y=5 4

When t=-1, y has a minimum value y=-1

y=sinx 2(sin x 2-cos x 2)=sin x 2-sinx 2*cos raid 2= (1-cosx)- sinx

½(sinx+cosx)+1/2

√2/2sin(x+π/4)+1/2

When sin(x+4)=1, y=sinx 2(sin x 2-cos x 2) has a minimum value: (-2+1) 2

When sin(x+4)=-1, y=sinx 2(sin x 2-cos x 2) has a maximum value: (2+1)2

The title should be: y=sinx 2(sin x 2-cos x 2).

Because the denominator can't be zero, x is not equal to 2k, there is no maximum and minimum value? （-1，1）

Solution: y=sin(x+6)+cosx

sinxcos(π/6)+cosxsin(π/6)+cosx=(√3/2)*sinx+(3/2)cosx=√3*[(1/2)*sinx+(√3/2)cosx]=√3*sin(x+π/3)

Because 0 x is 3 x + 3 4 3 - 3 2 sin(x+ 3) 1

Then when x+ 3 2, i.e. x 6, sin(x+ 3) 1, the function y has a maximum value of 3;

When x+3 4 3, i.e., x, sin(x+ 3) -3 2, the function y has a minimum value of 3 2.

x (0, ) then 0y=sinx+2 sinx is at 0, so it is sinx=1

Minimum=1+2 1=3

Solution: f(x)=2sinx(sinx+cosx)=2sinx+2sinxcosx

sin2x-(1-2sin²x)+1

sin2x-cos2x+1

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

x∈[0,π]

2x-π/4∈[-/4,7π/4]

The minimum value of sin(2x-4) [2 2,1] f(x) is: f(x)min=-1+1=0, and the maximum value is: f(x)min=2+1

f(x)=2sinx(sinx+cosx)=2sin²x+2sinxcosx

1-cos2x+sin2x

sin2x-cos2x+1

2sin(2x-π/4)+1

When 2x-4=2k+2,kz

i.e. x=k +3 8,k z, because x [0, ], so, when x=3 8

f(x) to achieve a maximum value of 2+1