I have questions about trigonometric functions, some trigonometric problems.

The question of trigonometric function is mainly the first big question in the college entrance examination, the overall difficulty is low, and the scoring rate is very high, so it is too simple to skip what the teacher said.

In fact, there are many ways to break through this kind of problem, because the question type of this kind of problem is very fixed, and the content is basically fixed in the induction formula of the trigonometric function part, and the difference angle formula, the doubling angle formula, the power of rise and fall and the most commonly investigated points in the application of the sine and cosine theorem.

The breakthrough method of this kind of problem is to start with the equation to find a breakthrough, usually the problem will give you an equation, then this equation usually has three situations, pure edge equation, pure angle equation and corner mixed equation, then the principle that must be grasped here is that the equation you can use must be either all edges, or all trigonometric relationships, not chaotic, so you only need to turn the equation of corner mixture into all edges or all angles.

Here are two tricks, use the sine theorem when replacing edges with angles.

When replacing angles with edges, the sine and cosine theorems are usually used.

After that, it's time to simplify it.

Another small note here is that in general, there are two very important implicit conditions, one is that the square of the sine plus the square of the cosine is equal to 1; The other is that the sum of the inner angles of the triangle is equal to 180, so it is common to replace 1 with the sum of two squares, and to cat the angle to -(a+b), etc.

With so many skills, this topic is not difficult, with this topic you can get about 110, it means that you still have some math skills, come on! I hope it can help you, ps: I'm tired, purely hand-hitting, forgive me for the shortcomings.

Is that okay?

Update 1:That's right).

as follows: Diagram Orange Tablets Reference: **Reference:

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**Reference: 2009-07-08 19:33:

48 Splend: 2009-07-08 19:33:

56 Supplement:

1.Derived from the universal formula.

sin2α=2tanα/[1+(tanα)^2]=-4/5

cos2 =[1-(tan ) 2] [1+(tan ) 2]=3 bright luck 5

sinα)^2=(1-cos2α)/2=1/5

sin cos =sin2 2=-2 Jing vertical beam 5

Original = 1 5-2 5 = -1 5

Answer: C2Because the fiber lift is a, b, c are the inner angles of the triangle, so a+b+c=

Derived from the induction formula.

cos(a+b)=-cosc

sin(a+b)=sinc

tan(a+b)=-tanc

sin[(a+b)/2]=cos(c/2)

Answer: B3Because y=3cos(2x+) the image is symmetrical with respect to (4 3,0).

So 3cos(2*4 3+)0, i.e. cos(+8 3)=0

So +8 3= 2+k* (k z).

k*π-13π/6

The minimum value is 6

Answer: A4Let f(x)=3-2(sinx) 2

Because f(-x)=3-2(sin-x) 2=3-2(-sinx) 2=3-2(sinx) 2=f(x).

So f(x) is an even function.

f(x)=3-2(sinx)^2=3-(1-cos2x)=2+cos2x

Because the period of cos2x is

So f(x) is an even function of t=.

Answer: Because the monotonically decreasing interval of cosx is [k, k+

So the monotonically decreasing interval of cos(2x-5) is [ 10+k 2,3 5+k 2].

Answer: [ 10+k 2,3 5+k 2].

6.The title is confusing and unclear.

Please will. Under the root number. Replace with.

Then check the question for ambiguity so that it can be resolved.

7.Because tanx >0, the terminal edge of the angle x is in the first or third quadrant.

According to sinx+cosx>0, sinx>0 or cosx>0 is obtained

If sinx > 0, the terminal edge of the angle x is in the first quadrant because tanx > 0.

If cosx > 0, the terminal edge of the angle x is in the first quadrant because tanx > 0.

In summary, the terminal edge of the angle x is in the first quadrant.

Here's an example.

Find the subtraction interval of the function y=2sin(3-2x).

Analysis] The monotonic interval of y=2sin(3-2x) can be obtained by first turning y=2sin(3-2x) into y=-2sin(2x-3), and then substituting 2x-3 as a whole into the corresponding monotonic interval of y=sinx.

Solution] Convert y=2sin( 3-2x) to y=-2sin(2x- 3), and find the decreasing interval of y=2sin(3-2x), that is, find the increasing interval of y=sin(2x- 3).

By 2k - 2 2x - 3 2k + 2, k z

Get k - 12 x k +5 12, k z

The subtraction interval of the function y=2sin( 3-2x) is [k - 12, k +5 12], k z

Note] In this problem, if you directly derive the range of x from 2k + 2 3-2x 2k + 3 2,k z, you get the monotonicity wrong. The reason for the mistake is to ignore the fact that y=2sin(3-2x) is essentially a composite of y=2sinx and y= 3-2x (this is a one-time function, decreasing in the defined domain), and the problem should be solved according to the monotonicity of the composite function, according to the principle of "same increase and different decrease".

When y=asin(wx+) is used to find the monotonic interval, it is better to be a positive number in front of x, and if it is not a positive number, it should be replaced with a positive number by the induction formula and then solved by overall substitution, so that it is not easy to make mistakes.

Using the sine theorem: a sina = b sinb

A=2BSINA, SINA=2sinBSINA

sinb = 1 2, b = 6, then c = 5 6-a

cosa+sinc=cosa+sin(5π/6-a)=cosa+sin5π/6cosa-sinacos5π/6=cosa+(1/2)cosa-sina×(-3/2)

√3/2)sina+(3/2)cosa=√3[sina×(1/2)+(3/2)cosa]=√3(sinacosπ/3+sinπ/3cosa)

3sin(a+π/3)

ABC is an acute triangle, A< 2, C=5 6-A< 2

32 31 2 3 2< 3sin(a+3)<3 2, i.e. 3 2 ranges (3 2, 3 2).

The sine and cosine theorem is a good solution to the relationship between angles and edges.

Sine theorem: a sina = b sinb

a=2bsina,sina=2sinbsina

sinb = 1 2, b = 6, then c = 5 6-a

cosa+sinc=cosa+sin(5π/6-a)=cosa+sin5π/6cosa-sinacos5π/6=cosa+(1/2)cosa-sina×(-3/2)

√3/2)sina+(3/2)cosa=√3[sina×(1/2)+(3/2)cosa]=√3(sinacosπ/3+sinπ/3cosa)

3sin(a+π/3)

ABC is an acute triangle, A< 2, C=5 6-A< 2

32 31 2 3 2< 3sin(a+3)<3 2, i.e. 3 2 ranges (3 2, 3 2).

No, there is no stipulation that f must be preceded by a positive number, any number is fine, anyway, this function is a periodic function.

If you have any doubts about trigonometric functions, then the above are the basics, and you can learn them again when you don't have a choice.

Trigonometric function is defined on the basis of a right-angled triangle, let me first say what it means, sin and cos will be followed by an angle, such as sin, cosx, then when you draw a right-angled triangle with one of the angles of or x in a Cartesian coordinate system or plane, then it will form a relationship about or the opposite side of x, the adjacent edge, the hypotenuse, where the opposite edge refers to the right-angled edge that is not in contact with the angle, and the adjacent edge refers to the right-angled edge next to the angle. An hypotenuse is an edge that is not in contact with the right angle of the right triangle.

In addition, regarding your first two questions, the functions of sin and cos do not completely depend on the planar Cartesian coordinate system, but they need to be analyzed through it; Not by the definition of the trigonometric function y=f(x), unless you turn xy up, like x=cosy.

Note: If you are good at math in junior high school, this is really not difficult, at least in terms of understanding.

Trigonometric functions (e.g., some polygons) can be calculated without a Cartesian coordinate system: y corresponds to the "opposite edge" in junior high school, x corresponds to the "adjacent edge" in junior high school, and the radius r of the unit circle is the hypotenuse (because any angle starts with the x-axis).

If you want to calculate, you can calculate it in **, refer to the first article. And if the angle is constant, the value of the trigonometric function does not change due to the change of the graph.

Additional: The definition of trigonometric functions in a unit circle is simple enough, sin = y r, a total of 2 English letters, 1 Greek letter, 3 mathematical symbols, what do you think is "simpler"?

Not necessarily, the Cartesian coordinate system is just an analysis, and x or y is just a letter that is used to distinguish between them.

Question 1:

abs(x) represents the absolute value of x, then 1 2 cos 3 4, can be written as 2 2 abs(cosx) 3 2 , using the knowledge of absolute values, you know that cosx cannot get the value between (- 2 2, 2 2).

Question 2: The trigonometric function cosx is generally considered to be defined by [-, then obviously 7 6 should be denoted as -5 6, and then the trigonometric function in the third quadrant image can be used to know that -3 4 and it is a pair of values, and so on.

Question 3: First, use 2k to write out the 4 intervals of the result, and then find that [2k -5 6, 2k -3 4] and [2k + 6, 2k + 4] are actually a pair, and they are exactly turned on the image, so they can be rewritten as [k + 6, k + 4]; Another analogy,

Question 1:

1/2≤cos²≤3/4

In addition to understanding these two.

3 2 cosx - 2 2 or 2 2 cosx 3 2 why can't this - 3 2 cosx 3 2 this is the basic solution inequality group!! The selection of the solution "Draw the number line" helps you to accurately determine the solution of the group of inequalities!!

I see for myself that your problem is that you don't have a conclusive solution to the region. Reason: How exactly do you determine the problem for the solution of the group of inequalities, how do you determine the solution for the domain of trigonometric functions and how they multiplicity. It is recommended that you take a look at the four compulsory courses!!

The mistake is that I haven't seen it clearly! The certainty of the solution makes you so entangled!

It is known that the domain of the function y=f(x) is [0,1 4] to find the domain of f(cos -1 2).

There are 0<=cos x-1 2<=1 41 2<=cos x<=3 4

1/2<=(cos2x+1)/2<=3/41<=cos2x+1<=3/2

0<=cos2x<=1/2

2kπ-π/3<=2x<=2kπ+π/3∴kπ-π/6<=x<=kπ+π/6

Trigonometric functions use the symmetry of a unit circle and the symmetry of a coordinate system. The main thing is to learn to transform and abstract and substantiate problems.

x is in the third quadrant.

So x<0, y<0

And r is always greater than 0.

So by y r=-1 2

You might as well set y=-1 and r=2

The size of another corner.

It is seen and x is half axis, and rotates counterclockwise.