Quick problem solving skills of functions. It's urgent ! Master, coming!

If you want to do it quickly, do more questions and remember them. The effects are obvious, but you can forget them quickly. There are not many questions in high school, so it will be easier if you have a good memory.

On several problem-solving skills of trigonometric function

I have accumulated some skills, experience and experience in solving problems in the face of trigonometric function teaching in vocational schools for more than ten years. Let's try to discuss:

First, about the promotion and application of relationships:

1, because it is known, it will be introduced, for example:

Example 1 known.

Analysis: Due to

Among them, as long as it is known, it can be solved. This problem is a typical problem of knowing the crime and seeking the crime.

Solution:

Therefore:

2. Application of the relationship between 2.tg +ctg and sin cos, SIN COS;

Because tg +ctg =

Therefore, knowing one of tg +ctg and sin cos, we can deduce the values of other formulas.

Example 2 If sin +cos =m2 and tg +ctg =n, the relationship between m2 and N is ().

m2=n B.m2= C. D。

Analysis: observe the relationship between sin +cos and sin cos;

Sine cos =

And:

So, choose B.

Example 3 is known: tg +ctg =4, then the value of sin2 is ().

A.B. C. D。

Analysis: tg +ctg =

So: the answer is a.

Example 4 is known: tg +ctg =2, find.

Analysis: As can be seen from the above example, as long as the formula containing sin cos or SIN COS can be written, it can be calculated according to the known tg +ctg. Because tg +ctg =

, just turn this problem into a formula containing sin cos:

Solution: = +2 sin2 cos2 -2 sin2 cos2

=(sin2 +cos2 )- 2 sin2 cos2

= 1-2 (sine and cosine) 2

= 1-

From the above example, we can draw the following conclusion: since sin cos and tg +ctg can be converted to each other, knowing one will make us know the other two. This property is suitable for the calculation of trigonometric formulas containing this trinomial. But one thing should be noted; If we find the included formula by knowing sin cos, we must discuss its quadrant to get the positive and negative sign of the result. This is because () 2 =12 sincos can only be found by root operation.

Second, about the application of "underpinning" method:

In the simplified calculation or proof of trigonometric function, it is often necessary to add denominator to the formula, which is often used in the reciprocity of formulas containing tg (or ctg) and sin (or cos). This method of adding letters is called the "bottom" method. The method is as follows:

Example 5: The value of tg =3 is known.

Analysis: Because there is a denominator cos, the original numerator and denominator can be divided by cos to "create" tg, which is the bottom: COS;

Solution: Because tg =3

Therefore, the original formula =

Example 6 is known: ctg = -3, find sin cos -cos2 =?

Analysis: Because of this, the formula will become an inclusive form, but this problem is different from Example 4. The formula itself has no denominator. In order to make the denominator appear in the original formula first, the formula: and bottoming methods are used to find its denominator, and then the numerator and denominator are divided by sin respectively to get ctg:

Solution:

Example 7 (95 National Adult College Entrance Examination for Science and Engineering Mathematics)

Settings,

Seek the value of

Analysis: This question is a typical formula for finding the tangent cotangent of sine function equation, and the "bottom finding method" should be used. Because of this, we should divide by the same number on both sides of the equation to get the denominator as the bottom.

Solution: Divide by both sides of the known equation:

"Underpinning" is suitable for the reciprocal calculation of formulas containing sine and cosine and formulas containing tangent and cotangent through the same angle. Because of the ratio relationship between tangent cotangent and sine cosine, their mutual transformation needs to "bottom out" By adding the denominator while keeping the formula values unchanged, they can be transformed into each other, so as to achieve the purpose of evaluating according to what is known. There are two main ways to increase the denominator: one is to use the denominator without changing the original formula value, and the other is to divide both sides of the equation by sine or cosine or their products to generate the denominator.

Third, the formal formula: and its application in solving the extreme value problem of trigonometric function;

We can get inspiration from the formula: the formula is somewhat similar to the above formula. If Part A and Part B are replaced by formulas containing sinA and cosA, then all formulas in shape can be replaced by included formulas, because-1≤ ≤ 1,

So we can consider using it to solve the extreme value problem, but we should pay attention to one thing: we can't directly regard A as Sina and B as cosA. For example, in the formula, sinA=3 and cosA=4 cannot be set. Consider:-1≤sinA≤ 1,-1 ≤ COSA ≤

Due to.

Therefore, we can set:, so it is:

∴

Whatever the value,-1 ≤ sin (a x) ≤ 1,

≤ ≤

Namely: ≤≤≤≤

Let's observe the application of this formula in solving practical extreme value problems:

Example 1( 1998 National Adult College Entrance Examination Mathematics Paper)

Find: The maximum value of the function is (AAAA)

A.B. C. D。

Analysis: Then try to turn it into an inclusive formula:

So:

Because here:

∴

Settings:

∴

No matter what value A-2x takes, there is-1≤sin(A-2x)≤ 1, so ≤≤≤≤

The maximum value of ∴ is, that is, the answer is A.

Example 2 (96 National Adult College Entrance Examination for Science and Engineering Mathematics)

In △ABC, it is known that: AB=2, BC= 1, CA=, take any point D, E and F on the sides of AB, BC and CA respectively, so that △DEF is a regular triangle, and note ∠FEC =∞α. Q: When sinα is taken, the side length of △EFD is the shortest. Find the shortest side length.

Analysis: First, it is known that △ABC is Rt△, where AB is the hypotenuse, diagonal ∠C is the right angle, and ∠B=

90-∠ A = 60。 Because this problem is to calculate the shortest side length of △DEF, we should set the side length of △DEF as, and list the unknown equations to solve. By observing △BDE, we know that ∠ B = 60, DE=, and then try to find out the other two quantities, and then we can list the equations according to the sine theorem, thus generating the equations about. In the figure, because EC=? Cosα, then BE=BC-EC= 1-? cosα.

And ∠ B+∠ BDE+∠ 1 = 180.

∠α+∠DEF+∠ 1 = 180∠BDE =∠α

∠B=60，∠DEF=60

∴ in △BDE, according to sine theorem:

Here, in order to have a minimum, the denominator must be: there is a maximum, observe:

∴

If:, then

Therefore:

The maximum value of ∴ is.

In other words, the minimum value is:

When the maximum value is 1,

∴

That is, △DEF has the shortest side length and the shortest side length is.

As can be seen from the above example, Shapu is suitable for calculating the extreme value of trigonometric functions. When calculating the extreme value, it has nothing to do with the addition and subtraction of the formula, but with the maximum value of; Where the maximum value is and the minimum value is. When calculating the extreme value application problem of trigonometric function, as long as the shape relationship is found, the related extreme value can be found according to the meaning of the problem.

Summary of solving problems of trigonometric function knowledge points

First look at the problem of "evaluation angle" and apply the "emerging" inductive formula.

One-step conversion to interval (-90? ,90? ) formula.

1 . sin(kπ+α)=(- 1)ksinα(k∈Z)； 2.cos(kπ+α)=(- 1)kcosα(k∈Z)；

3.tan(kπ+α)=(- 1)ktanα(k∈Z)； 4.cot(kπ+α)=(- 1)kcotα(k∈Z)。

Second, look at the problem of "sin α cos α" and use the triangle "gossip"

1 . sinα+cosα& gt； 0 (or

2.sinα-cosα& gt； 0 (or

3. The terminal edges of | sin α | > | cos α | α are in regions II and III;

4. The terminal edges of | sin α | <| cos α | ó α are in zone I and zone IV.

Third, look at the problem of "known 1 finding 5", do Rt△, and memorize the commonly used pythagorean numbers (3, 4, 5), (5, 12, 13), (7, 24, 25) by using pythagorean theorem.

Fourth, the problem of "cutting" is transformed into the problem of "string".

Verb (abbreviation of verb) "See Qi Sixian" = > "Change the chord into one": Given tanα, find the homogeneous formula of sinα and cosα. In some algebraic expressions, the denominator can be regarded as 1, which can be converted into sin2α+cos2α.

For intransitive verbs, refer to the "Square Difference of Sine or Angle" table, and enable the "Square Difference" formula:

1 . sin(α+β)sin(α-β)= sin 2α-sin 2β； 2.cos(α+β)cos(α-β)= cos2α-sin2β。

Seven, look at the problem of "sin α cosα and sinαcosα", and apply the square law:

(sin α cos α) 2 =1.2 sin α cos α =1sin2alpha, so

1. If sinα+cosα=t (and t2≤2), then 2sinα cosα = T2-1= sin2α;

2. If sinα-cosα=t, (and t2≤2), then 2sinαcosα= 1-t2=sin2α.

Eight. See the problem of "tanα+tanβ and tanαtanβ" and enable the deformation formula:

Tanα+Tanβ= Tan(α+β)( 1-TanαTanβ)。 Thinking: tanα-tanβ=?

Nine, look at the "symmetry" problem of trigonometric function, and enable the algebraic relationship of image features: (A≠0)

1. The images of function y=Asin(wx+φ) and function y=Acos(wx+φ) are symmetrical about the line passing through the maximum point and parallel to the Y axis, respectively;

2. Images with function y=Asin(wx+φ) and function y=Acos(wx+φ) are symmetrical about their middle zero centers respectively;

3. Similarly, the symmetry properties of function y=Atan(wx+φ) and function y=Acot(wx+φ) can also be obtained by images.

X. See the problem of finding the maximum value and range, and enable the boundedness or auxiliary angle formula:

1.|sinx|≤ 1，| cosx |≤ 1； 2.(asinx+bcosx)2 =(a2+B2)sin 2(x+φ)≤(a2+B2)；

3.asinx+bcosx=c has a solution if and only if a2+b2≥c2.

See "higher order", use power reduction, see "complex angle", use transformation.

1 . cos2x = 1-2 sin2x = 2 cos2x- 1。

2.2x =(x+y)+(x-y)； 2y =(x+y)-(x-y)； X-w=(x+y)-(y+w) and so on.

Angle function formula

Two-angle sum formula

sin(A+B)=sinAcosB+cosAsinB

sin(A-B)=sinAcosB-sinBcosA？