Senior 1 Math Proof Questions, 1 Senior 1 Math Proof Questions

sin(x+2x)(sinx) to the third power of +

cos(x+2x)(cosx).

sin2xcosx + cos2xsinx) (sinx) to the third power + cos2xcosx-sin2xsinx) (cosx).

2sinxcosxcosx+cos2xsinx) (sinx) to the third power of +

cos2xcosx-2sinxcosxsinx) (cosx).

2sinx 4th power cosx square + cos2xsinx 4th power + cos2xcosx 4th power - 2sinx square cosx 4th power.

2sinx to the fourth power cosx square + cos2x (sinx to the fourth power + cosx to the fourth power) - 2sinx square to the fourth power of cosx.

2sinx squared cosx square (sinx square - cosx squared) + cos2x (sinx quadratic + cosx quadratic) cos2x (sinx quadratic - 2sinx square cosx square + cosx quadratic).

cos2x (sinx squared - cosx squared) outside the square of parentheses.

cos2xcos2x squared.

cos2x).

It's best to grab a picture of this question, and I feel that the question can't be seen clearly, but the question is answered.

Proof: 1Since the function f(x) is symmetric with respect to x=a x=b, so f(x) = f(2a-x) and f(x) = f(2b-x) , so f(2a-x) = f(2b-x) = f(2a-x + (2b-2a)) i.e. f(x) = f(x+2b-2a), so f(x) ends with |2b-2a|is a periodic function of the period.

2.Because the function f(x) is symmetric with respect to a,0)(b,0), f(x)+f(2a-x)=0, f(x)+f(2b-x)=0

So f(2a-x) = f(2b-x) = f(2a-x + (2b-2a)) i.e. f(x) = f(x+2b-2a), so f(x) ends with |2b-2a|is a periodic function of the period.

3 Let (x,f(x)) be on the function f(x), then the function f(x) is about. The symmetry point of x=a is (2a-x,f(x).Also on the function f(x), so f(x)=f(2a-x) function f(x) with respect to (b,0) symmetry point is (2b-x,-f(x)) so:

f(x)=f(2b-x) because the point (2a-x,f(x).It is also on f(x), so its symmetry point about (b,0) is (2b-2a+x, -f(x)), so f(2b-2a+x)=-f(x) is obtained by f(2b-x)=f(2b-2a+x) so that t=2b-x, then x=2b-t, and f(2b-2a+x) is brought in f(t)=f(4b-2a-t), that is, f(x)=f(4b-2a-x) is combined to obtain f(2a-x))=f(4b-2a-x)=f(4b-4a+2a-x), i.e., f(x)=f(x+4b-4a), so f(x) is denoted by |4b-4a|is a periodic function of the period.

By the way, some of the above proves that some are relatively simple and do not understand how to communicate.

It's basically an example to do, and it's too cumbersome to write.

This is all a basic question, and the little brother doesn't give it).

There are examples of this in the book, and the teacher will talk about it in class.

Because x=a, x=b are axes of symmetry, there are f(a-x)=f(a+x), f(b-x)=f(b+x).

So f(x)=f(a-(a-x))=f(2a-x)=f(2a-x+b-b)=f(b-(b-2a+x))=f(b+(b-2a+x)))=f(x+2(b-a))).

So f(x) is a periodic function with a period of 2(b-a).

The distance from point o to the surface is half of the distance from point A to the surface, so find the distance from point A to the surface first. Find the midpoint E in B1D1, then the distance from A to the surface is the height of the CE side in the triangle ace, according to the geometric relationship, AC= 3, CE=(7) 2 (can be calculated in the triangle CB1D1), AE=CE. In the triangle ace, the height on the ac is 1, and the area of the triangle is, (3) 2, so the height on the CE side is (2 21) 7, then the distance from O to the plane CB1D1 is (21) 7

Let m = vector a·vector e

According to the title|a-te|^2≥|a-e|^2

a^2-2mt+t^2≥a^2-2m+1

t^2-2mt+2m-1≥0

For any real number, the above equation holds, and there is δ=(-2m) 2-4(2m-1) 0m 2-2m+1 0

m-1)^2≤0

So only m=1

i.e. vector a·vector e=1

So there is only e(a-e)=

i.e. vector e vector (a-e).

, b (-1,1) satisfies.

f(a),f(b),f[(a+b) (1+ab)] makes sense because f(x) = lg(1-x) (1+x).

So f(a)=lg(1-a) (1+a).

f(b)=lg(1-b)/(1+b)

f[(a+b)/(1+ab)]=lg[1-(a+b)/(1+ab)]/1+(a+b)/(1+ab)

f(a)+f(b)=lg(1-a)/(1+a)+lg(1-b)/(1+b)

lg[(1-a)/(1+a)(1-b)/(1+b)]=lg(1-a-b-ab)(1+a+b+ab)f[(a+b)/(1+ab)]

lg=lg[(1-ab-a-b)/(1+ab)]/[(1+a+b+ab)/(1+ab)}

lg(1-a-b-ab)(1+a+b+ab)=f(a)+f(b), i.e., f(a)+f(b)=f[(a+b) (1+ab)].

f(a)+f(b)=lg((1-a)(1-b)/(1+a)(1+b))

f(a+b 1+ab)=lg((1-(a+b) (1+ab)) 1+(a+b) (1+ab))))=lg(1-a-b+ab) (1+a+b+ab) (numerator and denominator multiplied by 1+ab at the same time.)

Then ab-a+1-b=a(b-1)-(b-1)=(a-1)(b-1)=(1-a)(1-b).

1+a+b+ab=a(b+1)+(b+1)=(a+1)(b+1)so lg(1-a-b+ab) (1+a+b+ab)=lg((1-a)(1-b) (1+a)(1+b)).

So f(a)+f(b)=f(a+b 1+ab).

Proof: Inversion method.

f(x)=lg[(1-x)/(1+x)]=lg[2/(1+x)-1]，f[(a+b)/(ab+1)]

lg[2(1+ab)/(1+ab+a+b)-1]=lg[(1-a)(1-b)/]

lg[(1-a)/(1+a)]+lg[(1-b)/(1+b)]=f(a)+f(b)

That is, f(a)+f(b)=f((a+b) 1+ab).

Because n>1, lgn>0, lg(n+1)>0, lg(n+2)>0;

To prove that the original inequality is true, it is only necessary to prove that lg(n+1) lgn>lg(n+2) lg(n+1);

That is: [lg(n+1)] 2>lgn lg(n+2)..

Because according to the mean inequality< [(lgn+lg(n+1)) 2] 2<[lg(n+1)] 2

Therefore, if (*) is true, all of the above steps can be reversed; So the original inequality holds,

Using the counter-argument method, assume that a, b, and c are all less than or equal to 0

then a+b+c should be <=0

a+b+c=(x-1) +y-1) +z-1) +3(x-1) ,y-1) ,z-1) both "=0, -3>0 so a+b+c is greater than 0."

contradicts assumptions.

So the assumption is wrong.

So it was proven.

Can you use QQ?,It's too troublesome to write like this.。

Let me think about it. Use the counter-argument and then add it up. The left is less than or equal to 0, while the right is greater than 0

Using the method of counter-evidence, hjm111 is very clear.