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I'll talk about the general idea, and if you don't understand, continue to ask.

1.Square the formula and then substitute the condition , and solve |b| = 3√22.In the same way, if the formula is squared, the required formula will be solved to obtain a·b + b·c + c·a = -13

3.Just get the angles into the familiar area and calculate the answer is 1 24With a 2x angle formula.

sin2α=2sinαcosα

cos2 =2cos 2( )1 can be squared into a perfect square of the formula in the root number, and then open and square according to the plus or minus values in it.

The result is: -2sin4

5.If you use the formula to open the angle, there will be (sin + cos), and then square it, and use sin 2( )cos 2( ) = 1 to find it.

sin2α = -7/ 8

1.Vector a*b=|a||b|*cos

2. (a+b+c )^2=0 . a*b=|a||b|*cos3.(No, ask other heroes).

4. sin(α+=sinαcosβ+cosαsinβ; 2) cos(α+=cosαcosβ-sinαsinβ;

Suggestion: The exercises are designed to strengthen the understanding and memorization of formulas, and in turn, to be familiar with the application of formulas, there is no need to do so many exercises.

Obviously, sin(2) >0

2θ∈(2kπ,π2kπ)

(kπ,π/2+kπ)，k∈z

Only d is not satisfied.

tan =2cos3 scrambled imitation 2sin3=cos3 sin3=1 tan3=cot3

Root retardation is based on cot(2)tan

cot(π/2-α)cot3

2-α=3+kπ

Altered limb 2-k -3 = k = 0

Let the c coordinate be (x,0).

k(ab)=(-3-1)/(5-2)=-4/3k(bc)=(0+3)/(x-5)

Since the angle abc = 90 degrees, there is k(ab)*k(bc)=-1, that is, there is: -4 3*3 [x-5]=-1

x-5=4x=9, i.e. c coordinate is (9,0), choose b

34. Take a=b=1 2 to get f(1)=2, take a=b=1 to get f(2)=4, so f(3)=f(2+1)=f(2)*f(1)=8.

35. Take x=y=1 to get f(1)=2f(1), so f(1)=0, take y=1 x to get f(1)=f(x)+f(1 x), so f(1 x)= -f(x), because f(2)=1, so f(4)=f(2*2)=2f(2)=2, then f(8)=f(2*4)=f(2)+f(4)=3, So f(64)=f(8*8)=2f(8)=6, so f(1 64)= -f(64)= -6.

31. The function image is symmetrical with respect to the straight line x=1, then f(x)=f(2-x), so f(4)=f(-2)= -(2) 2+1= -3. (First blank: -3).

And when x>1, 2-x<1 , so f(x)=f(2-x)= -(2-x) 2+1= -x 2+4x-3. (Second blank: -x 2+4x-3).

a=c, so open a power on both sides to get 2=c (1 a), the same way 5=c (1 b), multiply the two equations to get 10=c (1 a+1 b)=c 2, so c= 10. (Fill in: 10).

Note: Since c=2 a>0, it is rounded - 10).

The function image is symmetrical with respect to the line x=1, then f(x)=f(2-x) , so f(4)=f(-2)= -(2) 2+1= -3 and when x>1, 2-x<1 , so f(x)=f(2-x)= -(2-x) 2+1= -x 2+4x-3. (Second blank: -x 2+4x-3).

2 a=c, so open a to the power of both sides to get 2=c (1 a), the same way 5=c (1 b), multiply the two formulas, get 10=c (1 a+1 b)=c 2, so c= 10

13,4n+1, which can be obtained with slag auspicious sno+1-sn.

14. (-1,3) The determination theorem b 2-4ac<0 is obtained from the spike reed root determination theorem.