[q]

1 known cos (x+π/4) = 3/5, 17π/ 12.

2 known 4sin? x-6sinx-cos？ X+3cosx=0, find the value of cos2x-sin2x/(1-cos2x) (1-tan2x).

3 verification: cos? A+ sin? A+cos four times A-SIN four times a=4√2 times cos (π/4-A/2) times COS? a/2。

4.X∈(0, pai/2) is known.

Seek proof

sinx+tanx & gt； 2x

5. It is known that x>= y>= z>= 15 degrees, and x+y+z=90 degrees.

Find the maximum and minimum values of cosx*siny*cosz.

6. Given that the function f (x) = 2a (sinx) 2-2sqr (3) sinxcosx+b has a domain of [0, pi/2] and a range of [-5, -4], find constants A and B..

7. In △ ABC, ∠CAB=900, ∠C=300, AB= 1. Now P and Q start from point A at the same time, P moves at a constant speed along AC, and Q moves along AB and BC respectively, and the result reaches point C at the same time.

Find: 1) the speed ratio of point q to p.

2 let AP=x, S△APQ=y, and try to find the functional relationship between x and y when point q moves on the edge of BC.

3) When point Q moves on BC, what is the minimum value of S△ABQ+S△CPQ? (S△ABQ+ S△CPQ)

8. Let the domain of function f(x) be A={ 1, 2,3,4,5,6} and the range be B={ 1, 2,3}. For any A, b∈A, if a≤b, there must be F.

a， 1 1 b， 10 c，9 d，8。

9. If f(x) is satisfied: for any x, y, TF (x)+(1-t) f (y) > = f (tx+(1-t) y) holds, it is proved that [0, 1].

10.

The problem has been solved.

1. It is known that A, B and C are all positive angles.

And (Sina) 2+(sinb) 2+(sinc) 2 =1.

Verify a+b A+B+C & gt；; 90 degrees

[answer]

Let a, b and c be acute angles, then cos(A-B)>cos(A+B),

1=(sina)^2+(sinb)^2+(sinc)^2= 1-cos(a+b)cos(a-b)+(sinc)^2

& lt 1-[cos(A+B)]^2+(sinC)^2

That is cos (a+b)

therefore

a+B& gt； 90 degrees celsius

2. To make the function y = x2-2ax+ 1 have an inverse function on [1, 2], the value range of a is

[answer]

This function should correspond to [1, 2] as long as the axis of symmetry x=a is not in it.

That is a < = 1 or a & gt=2.

3. Given that sin α-sin β =- 1/3 and cos α-cos β = 1/2, find the value of cos(α-β).

[answer]

(sinα-sinβ)^2=sinα^2-2sinaαsinβ+sinβ^2= 1/9( 1)；

(cosα-cosβ)^2=cosα^2-2cosαcosβ+cosβ^2= 1/4②)，

( 1)+(2)= 2-2 cos(α-β)= 1/9+ 1/4。 Simplify!

4. Evaluation: sin faction /7*sin2 faction /7*sin3 faction /7 (2/7 and 3/7 respectively to avoid ambiguity)

[answer]

A=kπ/7(k=0, 1, 2, 3, 4, 5, 6) is the solution of the equation tan(7a)=0, let x=tana,

Because tan(4a)=-tan(3a), it's over.

X 6-21* x 4+35 * x 2-7 = 0, which is known from David's theorem.

Tan A * Tan (2a) * Tan (3a) * Tan (4a) * Tan (5a) * Tan (6a) =-7, while Tan (Ka) =-Tan [(7-k) A], k = 0, 1, 2, 3, 4, 4.

Tan (π/7) * Tan (2π/7) * Tan (3π/7) = SQRT (7), while COS (π/7) * COS (2π/7) * COS (3π/7) =1/8, so SIN (π/7) * COS =1/8.

5.Asinx+bcosx = 0，Asin2x+bcos2x-c = 0。

It is proved that 2 ABA+(b 2-a 2) b+(a 2+b 2) c = 0.

[answer]

Substitute Sinx=(-b/a)cosx with asinx+bcosx=0 into Asin2x+Bcos2x-C=0.

2a(-b/a)(cosx)^2+b[2(cosx)^2- 1]-c=0

That is, (cosx) 2 = a (b+c)/(2ab-2ba). If (B+C)/(2aB-2bA)=y, then (cosx) 2 = ay.

(sinx) 2 = radical sign (1-ay)

A 2 (sinx) 2+b 2 (cosx) 2+2 Absinxcosx = 0 If the squares of both sides asinx+bcosx=0.

That is, (1-ay) a+Yb 2 =-2bsinxcosx.

That is, a+(b 2-a 2) y =-2bsinxcosx, squared on both sides, you get.

a^2+2a(b^2-a^2)y+[b^4+a^4-2(ab)^2]y^2=4b^2( 1-ay)ay=4ayb^2-4(ab)^2y^2

That is, A2+2A (B2+A2) Y+[B4+A4+2 (AB) 2] Y 2 = 0.

[A+(A 2+B 2) y] A+(A 2+B 2) y where 2 = 0.

Substitute y=(B+C)/(2aB-2bA) into the above formula to simplify it.

6. Find the value of cos(π/7)-cos(2π/7)+cos(3π/7).

[answer]

Method 1:

Original formula = original formula * 2cos (π/14)/[2cos (π/14)]

= {[cos(3π/ 14)+cos(π/ 14)]-[cos(5π/ 14)+cos(3π/ 14)]+[cos(7π/ 14)+cos(5π/ 14)]}/[2cos(π/ 14)]

= 1/2

Method 2:

Let the original formula =u, because cos (π/7) * cos (2π/7) * cos (4π/7) = 2sin (π/7) * cos (2π/7) * cos (4π/7)/[2sin (π/7).

- 1/8 = cos(π/7)* cos(2π/7)* cos(4π/7)= cos(4π/7)*[cos(π/7)+cos(3π/7)]/2

=[cos(π/7)+cos(3π/7)+cos(5π/7)+cos(7π/7)]/4

=(u- 1)/4

So u= 1/2.

7. Given that angle A is acute, verify that 0 is less than sina+cosa (half of pi).

[answer]

Prove that a is an acute angle, then it is obvious that Sina+COSA >; 0。

Sina+cosa= (radical number 2)sin(a+45 degrees) = sin

8. A group

0； 10-x2 & gt； five

The function f(x) is an increasing function from 5 to positive infinity (open interval).

f( 10-x 1)>f( 10-x2)

That is, f (x 1) >: f(x2)

F(x) monotonically decreases from negative infinity to 5.

13.f(x) is a decreasing function defined on (-∞, 3), and the inequality f(a2-sinx)≤f(a+ 1+cos2x) holds for all ranges of values of x ∈ r.real number a.

[answer]

First, consider the domain: a 2-sinx.

Then consider monotonicity: a 2-sinx > =a+ 1+cos2x, namely a 2-a-1>; =(sinx+cos2x)max=9/8, and the solution is 1-sqrt (38)/2.

To sum up, the value range of a is-sqrt (2) < a <; 1

14. It is known that X 1 and X2 satisfy the equations (x+x = 3rd power of 10) and (lgX+X=3) respectively.

Then the value of X 1+X2 is _ _ _

[Answer] Let t= 10 be the x power, then x 1=lgt,

The first equation can be changed to t+lgt=3, which is exactly the same as the second equation.

So t=x2,

X 1+x2=lgt+t=3.

15.① For the function f(x) defined on r, if the real number x0 satisfies f(x0) = x0, x0 is called the function f(x).

Fix it. Let the quadratic function f(x) = x 2+MX-m+2 if f(x) has a fixed point on [0, +∞].

The value range of m is _ _ _ _ _ _ _

② function f (x) = ax 2+bx+c (a

The size relationship of is ()

a)f(0)& lt； f( 1)& lt； f(3)B)f( 1)& lt； f(3)& lt； f(0)

c)f(3)& lt； f(0)& lt； f( 1)D)f(0)& lt； f(3)& lt； f( 1)

③ Let the function f(x) be defined as (-∞, +∞), and the following function ⒈ y =-f (x) ⒉ y = f (x 2).

3.Y =-f (-x) 3。 Y = f (x)-f (-x) must be _ _ _ _ _ _ in odd function.

④ If the function f (x) =1+[m/(ax-1)] is odd function, then m = _ _ _ _ _ _ _

⑤ let a < b<c, then "function y = f(x) monotonically increases in the interval (a, b) and (b, c)" is.

() of "y=f(x) monotonically increases in the interval (a, c)"

A) necessary but not sufficient conditions b) sufficient and not necessary conditions

C) Necessary and Sufficient Condition D) is neither a sufficient condition nor a necessary condition.

[answer]

1.m is greater than double prime number 2-1 or less than negative double prime number 2-1.

2 . f(a+x)= f(a-x)= = = & gt； The straight line x=a is the symmetry axis of f(x), so choose C.

3. Option 4,

4. According to -f(x)=f(-x), divide by simplification to get M=2.

5.A

16. find cos 1. +cos2 .+cos3 .+ cos 179 .+ cos 180 .

( 1。 Refers to 1 degree)

[answer]

Cos(x)=-cos (u -x)

Such as cos 1. +cos 179 .=0

17. Given that the domain of f(x) is (0, +∞) and f(xy)=f(x)+f(y), the following expression is wrong ().

A.f( 1)=0

b.f(x^3)=3f(x)(x>； 0)

C.f(x)=0(x >0)

D.f (x under the radical sign) = (1/2) f (x) (x >; 0)

[answer]

f( 1 * 1)= f( 1)= f( 1)+f( 1)

So: 2f( 1)=f( 1), so f( 1)=0.

f(x * x * x)= f(x)+f(x * x)= f(x)+f(x)+f(x)= 3f(x)

Pay attention to x>0, x * x * x>0.

f(x)=f[sqrt(x)]+f[sqrt(x)]

C is wrong.

18.

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[answer]

② ④ Yes.

The range of f(x) is y ≠ 1, and the range of g (x) is y≠ 1, so p = q.

Issues to be discussed [/color]

Confusion of 1. inverse function

The teacher stressed that the domain of the inverse function must be found from the domain of the original function, otherwise it will make mistakes.

However, sometimes an example can find its domain directly from the inverse function. When will it be ready?

So confused!