Known functions f x 1 sinxcosx, g x cos x 12 10

Solution: (i)f(x)=1+sinxcosx=1+1 2 sin2x,g(x)=cos2(x+ 12 )=1 2 [1+cos(2x+ 6 )]x=x0 is an axis of symmetry of the function f(x) image, 2x0=k + 2 (k z), g(x0)=cos2(x0+ 12 )=1 2 [1+cos(2x0+ 6 )]=1 2 [1+cos(k +2 3 )].

When k is an even number, g(x0)=1 4; When k is odd, g(x0)=3 4

ii）h(x)=3 2 +1 4 sinωx+ 3 4 cosωx=1 2 sin(ωx+π 3 )+3 2

0, when x [-2 3 , 3 ], x+ 3 [2 3 + 3 , 3 + 3 ].

2 3 + 3 , 3 + 3 ] 2k - 2 ,2k + 2 ](k z), 2 3 + 3 2k - 2 3 + 3 2k + 2 , i.e. 3k+5 4 6k+1 2 , 0, 3k+5 4 0 6k+1 2 0 ,-1 12 k 5 12 ,k z, k=0, 1 2 , The maximum value is 1 2

Simplifying the above two functions, f(x)=1+sinxcosx=1 1 2sin2x g(x)=cos (x+ 12)=1 2 1 2cos 2x 6

2x0=π/2 ＋kπ g(x0)=﹙±√3﹚／2f（wx/2）+g（wx/2）＝3／2 ＋1／2 sin﹙wx＋π/3﹚

2 w﹚ －/3≥π/3

﹙2 w﹚－π/3≤-2π/3

W max is 3 2

1) Let x=x0 be an axis of symmetry on the image of the function y=f(x), and find the value of g(x0). kπ+π/4+π/12)]^2=[cos(kπ+π/3)]^2=(+-cosπ/3)^2=1/

Solution: (1) Because: f(x)=1+sinxcosx=1+12sin2x, its axis of symmetry: 2x=k + 2 x=k 2+ 4

And g(x)=cos2(x+ 12)=1+cos(2x+ 6)2 substitute x=k 2+ 4 into g(x)=1+cos(k + 2+ 6)21-sin 62=1-1212=14

2) Because: h(x)=f(x2)+g(x2)1+12sin x+1+cos(x+ 6)232+12sin x+12cos(x+ 6)32+12sin x+12(32cos x-12sin x)32+12(32cos x+12sin x)32+12cos(x-6).

When x [-2 3, 3], x-6 [-2 3- 6, 3- 6].

Because the function is an increasing function over the interval [-2, 3, 3].

So there must be -2 3- 6 - and 3- 6 0;

Solution: 54 and 12

Therefore, the maximum value is: 12

2) Find the monotonically increasing interval of f(x):

2) Find the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

ok(2) finds the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

2) Find the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

2) Find the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

2) Find the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

2) Find the monotonically increasing interval of f(x):

1) Find the minimum positive period of f(x):

17。The known function f(x)=1 2sinxcosx-3 4(cos 2x-sin 2x), x r

Solution: (1).

f(x)＝cos²(x＋π/12)＝1/2[1＋cos(2x＋π/6)]

x x0 is an axis of symmetry of the function y f(x) image.

2x0＋π/6＝kπ

i.e. 2x0 k 6 (k z).

g(x0)＝1＋1/2sin2x0＝1＋1/2sin(kπ－π/6)

When k is even, g(x0) 1 1 2sin( 6) 1 1 4 3 4

When k is odd, g(x0) 1 1 2sin( 6) 1 1 4 5 4

2)h(x)＝f(x)＋g(x)

1/2[1＋cos(2x＋π/6)]＋1＋1/2sin2x

1/2[cos(2x＋π/6)＋sin2x]＋3/2

1/2(√3/2•cos2x＋1/2sin2x)＋3/2

1/2sin(2x＋π/3)＋3/2

When 2k 2 x 3 2k 2

i.e. k 5 12 x k 12 (k z).

The function h(x) 1 2sin(2x 3) 3 2 is an increasing function.

Therefore, the monotonic increase interval of the function h(x) is [k 5 12,k 12](k z).

f(x)=sinxcosx- 3cos( +x)cosx sin 2x 3 3 2(x r), if the image of the function y=f(x) is translated by b= 4, 3 2 to obtain the image of the function y=g(x), that is, g(x sin 2x 3 4 3

sin 2x 12 3, g(x) on (0, 4) 12 2x 12 7 12, g(x) on (0, 4) maximum is 1 3

f(x)=sinxcosx- 3cos( +x)cosx(x r) can be converted to f(x)=(sin2x) 2+ 3((cos2x) 2+1)).

sin2x)/2+(√3cos2x)/2+√3=cosπ/3sin2x+sinπ/3cos2x+√3

sin(2x+3)+3The following transformation should be simple, can you do it yourself?!

1 2+1 2cos2x+1 2sin2x1 2+ 2 2( 2 2cos2x+ 2 2sin2x)1 2+ 2 2 (cos 4cos2x+sin 4sin2x)1 2+ 2 2cos(2x- 4).

1) When cos(2x- 4)=1, and ymax=1 2+ 2 22), x=a2+8 is substituted into f(x).

f(a 2 + 8) = 1 2 + 2 2cos[2(a 2 + 8)- 4] = 1, and the solution is a = 4

So s abc = 1 2 bcsina = 1 2 3 3 sin 4 = 9 2 4

Hope it helps.

There are some pickpocketing symbols that can't be played, forgive me!

1/2sin2x+1/2(cos2x+1)1/2(sin2x+cos2x)+1/2

Root number 2 2sin (2x+4) +1 2

So the period is ;

When the spring crack tease x [0, 2];

2x+π/4∈【π4，3π/4】

Then when 2x + 4 = 2, f(x) has a maximum value, which is .

root number 2 2 + 1 2;

When 2x+4=3 4, f(x) has a minimum value, which is.

Negative root number 2 2 + 1 2

Solution: f(x)=cos x-2sinxcosx-sin x-2sinxcosx+(cos x-sin x)-sin2x+cos2x

2*sin(2x-π/4)

1) The image is not convenient to make here, you can draw it yourself, it has been simplified to the general style, and it is easy to draw.

The interval is incomplete, is it [- 2,0]? Yes, the stupid words are solved as follows, if not, the method is also like Hu Ling, don't understand and ask again!

2≤x≤02*2-π/4≤2x-π/4≤-π4

Namely. 5π/4≤2x-π/4≤-π4

When 2x- band segment 4=- 2, sin(2x- 4)-1 is the minimum value, then the function obtains the maximum value - 2*(-1)= 2, when 2x- 4=-5 4, sin(2x- 4) 2 2 is the maximum value, and the function obtains the minimum value - 2*( 2 2)=-1 so the maximum value of the function f(x) on the interval [- 2,0] is 2, and the minimum value is -1

(1) Let x=xo be an axis of symmetry of the function y=f(x) image, and find g(x).

Since f(x)=1+(1 2)sin2x, for a sinusoidal function, the function f(x) takes the maximum or minimum value when x=xo is the axis of symmetry. That is: sin2x=1, or sin2x=-1

So, 2x=2xo=k +( 2)(k z).

Then, g(x)=[cos(2x+(6))+1] 2=[cos(k+(2)+(6))+1] 2

cos(kπ+(2π/3))+1]/2

When k is even, g(x)=[cos(2 3)+1] 2=[(-1 2)+1] 2=1 4

When k is odd, g(x)=[cos(5 3)+1] 2=[(1 2)+1] 2=3 4

2) Find h(x)=f(wx 2)+g(wx 2)(w0) in the interval [-2 3, 3] which is the maximum value of w of the increasing function.

As we can see from the above, f(x)=1+(1 2)sin2x,g(x)=[cos(2x+(6))+1] 2

So, f(wx 2) = 1 + (1 2) sin(2*wx 2) = 1 + (1 2) sin(wz).

g(wx/2)=[cos(2*wx/2+(π/6))+1]/2=[cos(wx+(π/6))+1]/2

So: f(wx 2)+g(wx 2)=1+(1 2)sin(wx)+(1 2)cos(wx+(6))+1 2).

3/2)+(1/2)[sin(wx)+cos(wx+(π/6))]

3/2)+(1/2)[sin(wx)+cos(wx)*cos(π/6)-sin(wx)*sin(π/6)]

3/2)+(1/2)[sin(wx)+cos(wx)*(3/2)]-sin(wx)*(1/2)]

3/2)+(1/2)[(1/2)sin(wx)+(3/2)cos(wx)]

3/2)+(1/2)sin[(wx)+(/3)]

3/2)+(1/2)sin[w(x+(π/3w))]

then its period is t=2 w

The length of the interval [-2 3, 3] is (3)-(2 3)=

To ensure that it is an increasing function on [-2 3, 3], then:

3w 2 3, and t 4 = (2w).

That is, the maximum value of w is 1 2