Urgently ask a few difficult math problems in the first year of high school, and if you can t, don t

Solution: ((1-sinx) (1+cosx).

(1-sinx)(1-cosx)/(1-(cosx)^2

1/sinx*(√1-sinx)(1-cosx))

1/sinx*(√1-2sinx/2cosx/2)(1-1+2(sinx/2)^2)=1/sinx*(√cosx/2-sinx/2)^2)(2(sinx/2)^2)=1/sinx*|cosx/2-sinx/2|*|sinx/2)|*2

1/sinx*|sinx/2cosx/2-(sinx/2)^2-1/2+1/2|√2

sinx+cosx-1)/√2sinx

1-cosx)/(1+cosx))=√(1-cosx)^2/(1-(cosx)^2)

1/sinx*(1-cosx)

Substituting: Original =

sinxcosx+(cosx)^2-1)/√2sinx)+1-cosx

sin(1999π+q)=sin(1998π+q+π)

sin(q+π)=-sinq

sin(2000π+q)=sinq

Because. f(1999)=4-(asinq+bcosp)=3

So. f(2000)=4+(asinq+bcosp)=5

If it is a series of equal differences, there is.

a6=a5+a7=3

a4+a8=3

a3+a9=3

a2+a10=3

a1+a11=3

So. s11=

Solution: 1) Is there a mistake in the title of your first question?

2) When x=1999, bring it into the original equation and get:

f(x)=asin(πx+q)+bcos(πx+p)+4-a-b+4=3

a+b=1 When x = 2000, bring it into the original equation and get :

f(2000)=a+b+4=5

3) This column is an equal difference series, which can be known according to the nature of the equal difference series:

AM+AN=AP+AQ (where M+N=P+Q), so A5+A7=A1+A11

a11+a1)*11 2=3*11 2=so n=11

I will, but I don't do it, I'll come to crap. Ha ha.

Will you really give points for doing it?

I still won't do it!

You copied the wrong question in the first question?!~

The first question must have been copied incorrectly.

Let sum be two equal difference series, denoted cn=max(n=1,2,3,...where max represents x1, x2 ,..., xs is the largest of the s numbers.

1) If an=n, bn=2n 1, find the values of c1, c2, c3, and prove that they are equal difference series;

2) Prove: or for any positive number m, there is a positive integer m, when n m, cn n>m; Or there is a positive integer m, such that cm, cm+1, cm+2 ,...is a series of equal differences.

Xueba is in the ring. Like one