It is known that the coordinates of A, B, and C are A3,0,B0,3, Ccosa, sina, and a 2,3,2 .

1) Vector oc=(cosa,sina), vector ab=(-3,3).

Vector oc Vector ab, cosa (-3)=sina 3

sina+cosa=0，∴√2*(√2/2*cosa+√2/2*sina)=0

i.e. 2*sin(a+ 4)=0, sin(a+ 4)=0

and a ( 2, 3 2), a + 4 (3 , 4, 7 4).

a+π/4=π，∴a=3π/4

2) Vector ac=(cosa-3,sina), vector bc=(cosa,sina-3).

Vector ac vector bc, vector ac*vector bc=0

cosa(cosa-3)+sina(sina-3)=0

That is, cos a-3cosa+sin a-3sina=0

That is, 1-3 (sina + cosa) = 0, sina + cosa = 1 3

sina + cosa) = 1 9, that is, sin a + 2 sinacosa + cos a = 1 9

1+sin2a=1/9，∴sin2a=-8/9

Original = [1+ 2*( 2 2*sin2a- 2 2*cos2a)] (1+tana).

1+sin2a-cos2a)/[1+(sina/cosa)]

2sin²a+2sinacosa)/[(sina+cosa)/cosa]

2sina(sina+cosa)]/[(sina+cosa)/cosa]

2sinacosasin2a look.

1)ac=(cosa-3,sina) bc=(cosa,sina-3)

Because |Vector ac|=|Vector bc

So (cosa-3) 2+sin 2a=cos 2a+(sina-3) 2

cos^2a-6cosa+9+sin^2a=cos^2a+sin^2a-6sina+9

Sorted out sina=cosa a=5 4

2) (2sin 2a+sin2a) (1+tana) finishing = [2sina(sina + cosa)] [(sina + cosa) cosa].

sin2a vector acVector bc=-1

cosa(cosa-3)+sina(sina-3)=-1cosa+sina=2/3

squared 1+sin2a=4 9

sin2a=-5/9

So the original = -5 9

First question:

From the coordinates of a, b, and c, we obtain: vector ac (cos 3, sin), vector bc (cos, sin 3), vector ac cos 3) 2 (sin ) 2 ,

Vector bc cos ) 2 (sin 3) 2 , according to the title, there are: vector ac vector bc , (cos 3) 2 (sin ) 2 (cos ) 2 (sin ) 2 (sin 3) 2, (cos ) 2 6cos 9 (sin ) 2 (cos ) 2 (sin ) 2 6sin 9, cos sin, again 90° 270°, 225°.

Second question:

Vector AC·vector BC 1, (cos 3)cos sin (sin 3) 1,(cos ) 2 3cos (sin ) 2 3sin 1,1 3(cos sin) 1, cos sin 2 3, (cos sin) 2 4 9,(cos ) 2 2cos sin (sin ) 2 4 9,1 sin2 4 9, sin2 5 9.

(1)ac=(cosa-3,sina) bc=(cosa,sina-3)

Because |Vector ac|=|Vector bc

So (cosa-3) 2+sin 2a=cos 2a+(sina-3) 2

cos^2a-6cosa+9+sin^2a=cos^2a+sin^2a-6sina+9

Sorted out sina=cosa a=5 4

2) （2sin^2a+sin2a)/(1+tana)

Finishing = [2sina(sina+cosa)] [(sina+cosa) cosa]

sin2a vector acVector bc=-1

cosa(cosa-3)+sina(sina-3)=-1

cosa+sina=2/3

squared 1+sin2a=4 9

sin2a=-5/9

So the original = -5 9

Solution: vector ac=(sin -3,cos), vector bc=(sin, cos -3).

Vector ac|=√sinα-3)^2+(cosα)^2]=√sinα)^2-6sinα+9+(cosα)^2]=√10-6sinα)

Vector bc=√sinα)^2+(cosα-3)^2]=√sinα)^2+(cosα)^2-6cosα+9]=√10-6cosα)

Vector ac|=|Vector bc

(10-6sinα)=10-6cosα)

-4= , then =5 4

2) Vector ac* vector bc

sinα-3)*sinα+cosα* cosα-3)

sinα)^2-3sinα+(cosα)^2-3cosα

1-3(sinα+cosα)

Vector ac*vector bc=-1

1-3(sin +cos) = 1, solution: sin +cos = 2 3

sinα+cosα)^2=(sinα)^2+2sinαcosα+(cosα)^2=1+2sinαcosα=4/9

2sinαcosα=-5/9.

2(sinα)^2+sin(2α)]1+tanα)

2(sinα)^2+2sinαcosα]/cosα/cosα+sinα/cosα)

2(sinα)^2+2sinαcosα]/sinα+cosα)/cosα]

cosα*[2(sinα)^2+2sinαcosα]/sinα+cosα)

2(sinα)^2*cosα+2sinα*(cosα)^2]/(sinα+cosα)

2sinαcosα(sinα+cosα)/sinα+cosα)

2sinαcosα

I don't know if it's right or not, if there is a mistake, please point it out, thank you!

I'm so sorry, I almost forgot about it.

Hello:(1) |Vector ac|=|Vector cb|

c is on the perpendicular line of ab, let the midpoint of ab d(3 2,3 2) dc vector ab vector = 0

cosα-3/2,sinα-3/2)(-3,3)=0∴sinα=cosα

2) Vector ac*vector bc=-1

cosα-3)cosα+(sinα-3)sinα=-1∴cosα+sinα=2/3

Landlord, is there any parentheses in your 2sin 2a+sin2a 1+tana?

Hello! Vector AC = (cosa-3,sina) vector BC = (cosa,sina-3) vector AC·vector BC

cos²a - cosa + sin²a - 3sina= 1 - 3(sina+cosa)

sina + cosa = 2/3

tan(a/2) +1/tan(a/2)= sin(a/2) / cos(a/2) +cos(a/2) / sin(a/2)

sin²(a/2) +cos²(a/2) ]/ [ sin(a/2) cos(a/2) ]

1 / ( 1/2 sina)

2 / sina

Original = (2sin A + sin2a) 2 sina= (2sin A + 2sinacosa) *2 sina = 4sina + 4cosa

1、|ac|=|bc|i.e. (cos -3) + sin cos sin -3), the solution gives sin =cos, i.e. =5 4

2. The absolute value of the vector will not be equal to -1, which is wrong.

1) Using the image method, c is the left semicircle of the unit circle, and the solution is =5 4

2) I really don't know what he's talking about.

Solution: The combination of numbers and shapes can be known.

c (cosa, sina) is a point on the unit circle.

By |ac|=|bc|Know.

Point c (cosa, sina) must be on the angular bisector of the first and third quadrants, cosa=sina

tana=1.Combined with a ( 2, 3 2) a = 5 4

(1) Actually, the title says that the modulus of the two vectors is equal, not the absolute vector ac=(cosa-3,sina).

Vector bc = (cosa, sina-3).

modulo of vector ac = root number (10-6cosa).

Modulo of vector bc = root number (10-6sina).

So cosa=sina

Because a belongs to (2, 3, 2).

a=5π/4

2) The question vector ac*bc=(cosa) 2-3cosa+(sina) 2-3sina=-1

cosa+sina=2/3

What does the title mean? The value of 2sin squared a+sin2a 1+tana.

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