For the problem of co angular trigonometric functions f cosx x 2, find f cos 4 3

Summary. When solving a math problem, a certain formula is treated as a whole and a variable is substituted for it, so that the problem is simplified, which is called the commutation method. The commutation method is also known as the auxiliary element method and the variable substitution method.

The essence of the exchange element is transformation, the key is the construction element or the set element, the theoretical basis is the equivalent substitution, the purpose is to introduce new variables, the scattered conditions are linked, the implicit conditions are revealed, the conditions are linked to the conclusion, the unfamiliar form is changed into a familiar form, the complex calculation and deduction are simplified, and the non-standard problems are standardized. It has a wide range of applications in the study of equations, inequalities, functions, sequences, trigonometry and other problems, and is a standing method for mathematics in the college entrance examination. This paper takes solving the derivative problem as an example to illustrate the role of the commutation method in optimizing the problem solving process or finding a way to solve the problem.

Knowing that f(cos x)=tan x+3, the problem of finding f(x) can be calculated using the commutation method.

Let cos x=t

then tan x = (1-cos x) cos x then and left = (1-t ) t

then the left = (1-t) t 3

Both: f(x)=(1-x) x 3

Then f(t)=(1-t) t 3.

Turning t into x is the answer.

When solving a math problem, a certain formula is treated as a whole and a variable is substituted for it, so that the problem is simplified, which is called the commutation method. The commutation method is also known as the auxiliary element method and the variable substitution method. The essence of the exchange element is transformation, the key is the construction element or the set element, the theoretical basis is the equivalent substitution, the purpose is to introduce new variables, the scattered conditions are linked, the implicit conditions are revealed, the conditions and conclusions are linked, the unfamiliar form is changed into a familiar form, the complex calculation and deduction are simplified, and the non-standard problems are standardized.

It has a wide range of applications in the study of equations, inequalities, functions, sequences, trigonometry and other problems, and is a standing method for mathematics in the college entrance examination. In this paper, we take solving the derivative problem as an example to illustrate the role of the slag commutation method in optimizing the problem solving process or finding a way to solve the problem. Good.

f(x)=-3cos2x-sin2x=-2[(√3/2)cos2x+(1/2)sin2x]=-2cos(2x-π/6)

The minimum reputation t=2 2=

x=π/12,f(π/12)=-2

x=π/3,f(π/3)=0

2≤f(x)≤0

3 x Jing Hu infiltrate 12 monotonically decreasing.

12 x 3 monotonous bright ridges increase.

Solution: (1) Simplification first.

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=sinxcosπ/6+sinπ/6cosx+sinxcosπ/6-sinπ/6cosx+cosx+a

3x sinx + cosx + a

2sin(x+π/6)+a

When sin(x+6)=1, f(x) is maximum, and the maximum value is 2+a 2+a=1

a=-12) from (1) to obtain f(x)=2sin(x+6)-1f(x)>=0, i.e., 2sin(x+ 6)-1>=0 sin(x+ 6)>=1 2

2k + 6<=x + 6<=2k +5 6 The solution yields (3) x [0, ]x+ 6 [ 6,7 6] The maximum value of f(x) is 1, and the minimum value is f( )=2sin(7 6)-1=-2

The value range is [-2,1].

Proof: 2f(x)=(4cos 4x—4cos x+1) (2tan( 4—x)cos ( 4-x))).

2cos^2(2x)-1]^2/(2sin（π/4—x）cos（π/4-x））

cos2x) 2 sin( 2-2x)=cos2x(2)f(x)=2 5, so cos2x=4 5 x (0, 2), so 2x (0, 2), so sin2x=3 5, tan2x=3 4 let tanx=m, then m>0, 2m (1-m 2)=3 4 solution gives m=1 3, so 3tanx=1

Take the known conditions, calculate the value of cosx-sinx, square it, and get the value of 2sinxcosx. Then use the relationship between the square of cosx-sinx and the square difference of cosx+sinx by 4sinxcosx, calculate the value of cosx+sinx, pay attention to the value range of x at this time, and the value of cosx+sinx should be negative! Simplify the formula to be sought, tanx is written as sinx cosx, the denominator is divided, and the numerator extracts the common factor, and what you get is the formula about cosx-sinx, sinxcosx, cosx+sinx, and substitute the value just now to come out, hehe.

a=pie 4+x can be judged to be the fourth quadrant, so sina = -4 5

tana=-4/3 tan(2a)=24/7=-cot(2x)

Result = 2tan 2x(1+tanx) [(1-tanx)(1+tan 2x)] = tan(2x)*tan(pai 4+x) = -7 24*(-4 3) = 7 18

f(x) is an even function.

f(x)=f(-x)

sin(-x+θ)cos(-x-θ)sin(x+θ)cos(x-θ)sin(x-θ)cos(x+θ)sin(x+θ)cos(x-θ)sin(x-θ)cos(x-θ)cos(x+θ)sin(x+θ)2sin(x-θ+4)=√2sin(x+θ-4)x-θ+4=x+θ-4 ①

or x- +4+x+ -4 = 2

Derived from = 4

Therefore, it is only true if x = 4.

So to sum up = 4

According to the definition of the even function, f(- f( ) belongs to (0, ), thus finding sin2 -cos2 =-1, i.e. the root number 2sin(2 - 4) = -1

The binding range is =3 4

I vaguely remember that there is a formula for the 2x angle, maybe it works, but sometimes this thing doesn't require a specific value. It's okay to turn it into a function.