I was in a hurry to ask for a math problem in my first year of high school!!!!!!!!

0< 0, sin >0, sin >0coss >0 ,cos >0, cos >0sin( + sin( +sin sin must be proved. sin(θ+sinβ>sin(θ+sinαsinθcosα+sinαcosθ)sinβ>(sinθcosβ+sinβcosθ)sinα

sinθsinβcosα+sinαsinβcosθ>sinαsinθcosβ+sinαsinβ*cosθ

Subtract sin sin cos on both sides

Get sin sin cos > sin sin cos both sides divided by sin

sinβcosα>sinαcosβ

sinβcosα-sinαcosβ>0

sin(β-0

sin( -0 certificate ended.

sin(θ+=sinθcosα+sinαcosθ

sin(θ+=sinθcosβ+cosθsinβ

So, to prove.

sin(θ+/sin(θ+=(sinθcosα+sinαcosθ)/(sinθcosβ+cosθsinβ)>sinα/sinβ

must be proved:

sinθcosα+sinαcosθ)*sinβ>(sinθcosβ+cosθsinβ)*sinα

That is, to prove: sin cos sin +sin cos sin > sin cos sin +cos sin sin

Both sides make an appointment with sin cos sin at the same time

So, that is, to prove.

sinθcosαsinβ>sinθcosβsinα

sinθcosαsinβ-sinθcosβsinα>0

sinθ(cosαsinβ-cosβsinα)>0

sinθsin(β-0

And because. f(x)=sinx increases monotonically on (0, 2), and 0< so sin >0 and sin( -0 is proved.

1 Because it is an odd function, f(0)=0, b=0

Substituting f(1)=1 2, we get 1 (1+a)=1 2, a=12 f(x)=x (x 2+1).

f(x)'=[(x^2+1)-2x^2]/(x^2+1)^2=(1-x^2)/(x^2+1)^2

Req f(x).'>0 gives x to belong to (-1,1);

f(x) is an increasing function over the interval (-1,1).

3 g(0)=3^0-0=1

g(1)=1/3-1/2=-1/6<0

Whereas g(x) is continuous, so there is a zero point on (0,1).

So the function g(x) has a zero point on (-.

3sinx-4cosx-k =0

Using the auxiliary angle formula, tidy up.

5sin(x+s) =k,s =arctan(-4/3)

sin(x+s) =k/5

sin(x+s) makes sense.

1≤k/5≤1

Solution, -5 k 5

Divide both sides of the equation by 5 to get 3 5*sinx-4 5*cosx=k 5;

Order: cosy=3 5, then: siny=4 5, have: sin(x-y)=k 5

It is obtained from -1<=sin(x-y)<=1, -1<=(k 5)<=1, -5<=k<=5

1.Solution: When x [0,2], the function obtains the maximum value when x = 2, that is, the original function increases monotonically on x [0,2].

The axis of symmetry of the original function is (2-2a) a

When a 0.

2-2a）/a≤0

a 1 when a 0.

2-2a) a 2, get a 1 2

a Empty set. In summary, a 1

2.Solution: f(x) is the odd function, -f(-1)=f(1)>1

i.e. f(-1)<-1

and the period t=3 of f(x).

f(-1)=f(-1+3)=f(2)<-1, i.e. (2a-3) (a+1)<-1

Add 1 to both sides of the unequal sign

Simplification, obtainable:

3a-2)/(a+1)<0

13.Solution: Let x=y=3

Then f(xy)=f(9)=f(3)+f(3)=2 The original inequality can be reduced to f(x)+f(x-8) f(9) The original function is an increasing function.

The original inequality can be reduced to x+x-8 9

Solution: x 17 2

4.Solution: Let f(x)=ax +bx+c

f（x+1）+f（x-1）=2x²-4x∴a(x+1)²+b(x+1)+c+a(x-1)²+b(x-1)+c=2x²-4x

2a=2,4a+2b=-4,2a+2c=0 is solved: a=1, b=-4, c=-1

f(x)=x²-4x-1

1,-4（a-1）/2a=2

4a+4=4a

a=-1 2, choose a

2, because it is an odd function, so -f(2)=f(-2)=f(1) 1 is: (2a+3) (a+1)<-1

Solution: -4 3

is an odd function defined on r with a period of 3.

f(2)=f(2-3)=f(-1)=-f(1）=（2a+3）/（a+1)，f(1)>1 -f(1)<-1

2a+3）/（a+1)<-1

4 30, f(x) is an increment x 2-8x<9 x 2-8x> 0-18 on the defined domain (0,+, 89 x 2-8x<0x<-1 or x>9 0< x<8x=0 f(x)+f(x-8)=f(8)4Let f(x)=ax 2+bx+c

f(x+1)=a(x+1)^2+b(x+1)+cf(x-1)=a(x-1)^2+b(x-1)+cf(x+1)+f(x-1)=2ax^2+2bx+2c+2a2a=2 a=1

2b=-4 b=-2

2c+2a=0

c=-1f(x)=x^2-2x-1

Let's talk about the idea: Establish the function f(x)=1 Sidka 5*(100-x)+2 (root number x).

Replacing it with the root number x with t = is a quadratic function hole early, and I have learned Nalufinch in junior high school. Find f(x)max=

I don't understand hi me.

1) Find the swing angle (radian) of the initial position

At the initial moment, t=0, bring in f(t), and get the swing angle of the initial position f(0)=1 2;

2) Find the frequency of the pendulum.

Frequency =2 2= ;

3) Find how long a single pendulum completes 5 complete swings (one reciprocating swing is called a complete swing) period t = 1, 5 complete swing time t'=5t=5/π。