Ladies and gentlemen, how to do this problem, seconds. 20

In short, starting from coscosx and sinsinx, the left is always between cos1 1, the right is always between 0 sin1, and cos1 >sin1, which is completed.

As for how to think,,, it is probably that after the first cos and sin, all x must be between -1 and 1, and no matter how cos is in this interval, it must be absolutely positive, so sinx is only between 0 1 and the ratio of the two, at this time cos is also in 0 1, (the cos operation is meaningless after the positive and negative pairs), so x only needs to be considered between 0 pi 2, and from the first layer onwards, only need to be considered between 0 1, and cos in this interval is always subtracted and sin increases.

Subsequent attempts to prove cosa>sin (1-a2), a [0,1], deformation and derivation were fruitless. Then try to increase and decrease the contraction, and easily prove that x>sinx>sinsinx>......But the test proved that coscos(x)>x was fruitless. Overturn and redo, continue the original idea and continue to try, and find that the first cos is 0 1, the second time is cos1 1, and the third time is cos1 coscos1......The maximum value of the previous layer will not be greater than 1, and the minimum value of the next layer will not be less than cos1; In the same way, if you look for sin, you will find that it is 0 1 after the first time, 0 sin1 after the second time, and 0 sinsin1 after the third time......From the previously known sinx>sin n(x), cos1>sin1 was tested, and the solution was successful.

It is a loose confusion condition with 2n, but it is actually a single n, and n>=2 is sufficient.

1) When sinx < 0, sinx0

Hence sin (sin....(sinx))<00

In this case, just compare sin(sinx) and cos(|cosx|) size.

cos(|cosx|)=sin(π/2 - cosx|Because |cosx|+sinx= 2 sin(x+ )2, so sinx 2- |cosx|<π/2 - cosx|< 2 So sin(sinx) has proved sin(sinx) so sin(sin(sinx)) sin(sin1) so sin(sin....).(sinx))

You have to use the inductive method to do this problem, prove: by sinx, cosx's minimum positive period is 2 factions (the input method can't be typed, sorry), just prove that when x belongs to [0,2 factions], make sin(sin(sin(sin....)sinx)) (2n in total) cosx)) (2n in total).

When x=0, sin0=0, cos0=1, the left side of the inequality = 0, the right side of the inequality is 0, and the inequality holds; When x=faction, sin faction=0, cos faction=-1, inequality left = 0, right "0", the inequality holds; When x=3 is 2, sin3 is 2=-1, cos3 is 2=0, the left side of the inequality is <0, and the right side is 0, and the inequality holds. When x belongs to (3 factions 2, 2 factions), the inequality is <0 on the left, 0 on the right, and the inequality holds; When x belongs to (faction, 3 faction 2), the inequality is <0 on the left, and 0 on the right, and the inequality holds; When x belongs to (0, pie 2), sinx + cosx = root number 2sin (x + pie 4) belongs to (1, root number 2], i.e. there is sinx + cosx "pie 2, i.e. 0 and so on sin(sin(..sin(x)) (2n sin) look, thank you.

From the view, it can be seen that the bottom surface of the triangle in the top view is 6 long, the ground height is 4, and according to the Pythagorean theorem, the waist length is 5, so the side area is equal to.

Solution: From the meaning of the title, it can be seen that the number of this series is divided by 1 and is divisible by 7.

So, there are a total of 119 steps on this ladder.

119 floors.

The 2nd floor cannot be completed, indicating that it is an odd number.

5 floors will have 4 floors left, indicating that the single digit is 9.

The 7th layer is just finished, indicating that it is divisible by 7.

7 7 = 49 does not match.

17 7 = 119 compliant.

o(∩_o

Is the landlord asking how many steps there are? You first see, each step has two orders, and there is one order left, which means that the order is singular, so find a multiple of 3 plus 2 is the number of a singular number. Again, adding two is singular, indicating that the multiple of the previous 3 must also be singular, then the final answer may be ·· 3n+2 (n is singular).

In fact, there is no exact answer to this question, 119, 329, 539, 749, 959, 1169 every 210 is satisfied.

Explanation: If you add 1 to the number of steps, you will find that every 2, 3, 5, and 6 steps can be completed.

The least common multiple of is 30

That is, the number of steps is a multiple of 30 minus 1

So it can only be ......

But only 119,329,539,749,959,1169 ...... divisible by 7

I guess the answer is 119, and I don't usually build that many stairs.

Solution: Let the original number be x, and it is known that the decimal point is moved to the right by two places, then: the new number is 100x

Also know: the new number is larger than the original.

Yes: 100x-x=

i.e.: 99x=

Then: x = A: The original number is.

Moving the decimal point to the right by two places expands the original number by a factor of 100.

A decreasing proportional series, [otherwise there can't be a "sum of all terms"], so -1 q 1 can use the range of quadratic functions.

The square is 1 and the circle is 2

The triangle is 5, and the figure is plus + in the square

The figure outside the square is minus-