Trigonometric function and maximum, a formula for finding the maximum value of trigonometric functio

Because 4 x 2, 2 2 sinx 1 sets t=sinx, 2 2 t 1

The original formula is y=2 t+t 2

Derivative y'=(-2 t )+1 2

2 2 t 1, so 1 2 t 1, 1 1 t 2, -4 -2 t -2, -7 2 (-2 t )+1 2 -3 2

It can be seen that the derivative is always less than 0, and the function decreases monotonically when it is 4 x 2, and the maximum value is obtained when x = 2 2.

Substituting calculates 2 ( 2 2) + (2 2) 2 = (9 2) 4, so the answer is c

Let t=sinx [ 2 2,1].

y=2/t+t/2

It is a checkmark function that decrements on (0,2) and increases on (2,+).

So, y=2 t+t 2 is decreasing on [ 2 2,1].

So when t=2 2, y has a maximum value =2 ( 2 2) + (2 2) 2=2 2+ 2 4=9 2 4 choose c

Because sinx [ 2 2 , 1].

sinx x >0, x sinx >0, then y=2 sinx+sinx 2=2 (2 sinx*sinx 2)=2

Derivatives are complex.

Or because sinx>0, the function has a minimum value (when x>0, f(x)=ax+b x has a minimum value (specifying a>0, b>0), that is, when x=sqrt(b a) (sqrt means finding the quadratic root)), that is, sinx=sqrt(2 (1 2))=2 has a minimum value y=2

2 sinx is seen as x then y=x+1 x sinx is between (root number 2 divided by 2,1) 2 sinx is in (2,2 root number 2), by monotonicity, choose c

In fact, there are six kinds of simplest trigonometric functions, y=sinx, y=cosx, y=, y=cscx, the first four are called basic elementary functions, y=tanx, y=cotx functions have no maximum, and for y=asin(bx+c)+m, (a does not equal 0, b does not equal 0), its maximum value is a+m, the other trigonometric functions are the most complex and difficult to solve, you need to find the derivative of the original function, make the derivative of the function 0, get the coordinates of the feature points, and finally make a comparison, the largest eigenvalue is the maximum value of the function. For multivariate functions, you must select the coordinate axis, see which axis is taken as the maximum, and then find its partial derivative for the axis, and apply the above method.

Finding the maximum value of a trigonometric function usually has the following types:Type 1: One-time homogeneous type.

Auxiliary angle formula, turn into an angle to find the maximum.

Type 2: Quadratic homogeneous type.

To reduce the power of the auxiliary angle, you need to use the power reduction formula and the auxiliary angle formula, and find the maximum value twice.

Type 3: Quadratic non-homogeneous.

Converted into the form of quadratic function, the formula is the maximum value, and the range needs to be paid attention to.

Type 4: Fractional type.

The inverse method takes advantage of the boundedness of trigonometric functions.

Type 5: Commutation method.

Pay attention to the range of the parameter t after the commutation, which is usually a quadratic function after the commutation, and the maximum value is found through the formula.

The issues to be aware of are:(1) Pay attention to the given interval of the question.

2) Pay attention to the equivalence of algebraic substitutions or trigonometric transformations.

3) Trigonometric formulas with parameters, the role of parameters should be emphasized, and it is likely to be discussed.

Trigonometric Function Definition:Trigonometric function is one of the basic elementary functions, which is a function in which the angle (the most commonly used radian system in mathematics, the same below) is the independent variable, and the angle corresponds to the coordinate of the final edge of any angle and the intersection point of the unit circle or its ratio as the dependent variable.

It can also be defined equivalently in terms of the length of the various line segments related to the unit circle. Trigonometric functions play an important role in the study of the properties of geometric shapes such as triangles and circles, and are also a fundamental mathematical tool for the study of periodic phenomena. In mathematical analysis, trigonometric functions are also defined as infinite series or solutions to specific differential equations, allowing their values to be extended to arbitrary real values, even complex values.

Both sinx and sin(2x-6) are forms of the trigonometric function f(x)=sin(x).

You can make t=2x- 6 then sin(2x- 6)=sin(t).

That is, to make sinx and sint have the same form.

t= 2 when sint i.e. sin(2x- 6) has a maximum value.

At this point 2x- 6=t= 2 so x= 3

Find the monotonic interval of the sint to get the interval about t.

Then according to t=2x-6, we can calculate the monotonic interval of sin(2x- 6) with respect to x.

sint t=Both sinx and sin(2x-6) are forms of the trigonometric function f(x)=sin(x).

You can make t=2x- 6 then sin(2x- 6)=sin(t).

That is, to make sinx and sint have the same form.

t= 2 when sint i.e. sin(2x- 6) has a maximum value.

At this point 2x- 6=t= 2 so x= 3

Find the monotonic interval of the sint to get the interval about t.

Then according to t=2x-6, we can calculate the monotonic interval of sin(2x- 6) with respect to x.

t=90 degrees to find the maximum point A.

Whether it's sinx or sin(2x-6).

They are all forms of trigonometric functions f(x)=sin(x), and you can make t=2x-6

then sin(2x- 6) = sin(t).

That is, to make sinx and sint have the same form.

t= 2.

sint i.e. sin(2x-6) has a maximum value.

At this point 2x- 6=t= 2

sox=π/3

Find the monotonic interval of the sint to get the interval about t.

Then according to t=2x-6, we can calculate the monotonic interval of sin(2x- 6) with respect to x.

Sintt=Both sinx and sin(2x-6) are forms of the trigonometric function f(x)=sin(x), and you can make t=2x- 6

then sin(2x- 6) = sin(t).

That is, to make sinx and sint have the same form.

t= 2.

sint i.e. sin(2x-6) has a maximum value.

At this point 2x- 6=t= 2

sox=π/3

Find the monotonic interval of the sint to get the interval about t.

Then according to t=2x-6, we can calculate the monotonic interval of sin(2x- 6) with respect to x.

t=90 degrees.

Find the maximum point.

Here's how:

Let 2x- 6=t.

The coefficient 2 of the original formula = 2sin(2x- 6)= does not affect his maximum point, so we can ignore it. I'm sure you should know the maximum particle of sint! T = 2 of course (in one cycle, of course).

And because 2x- 6=t, this brings out the equation you smell: 2x- 6= 2. The cycle is needless to explain.

In trigonometric functions, the independent variable is the angle, and the variable is the ratio, i.e., the function, ,

Tangent, the sine is getting bigger and bigger、、、 the cosine is getting smaller and smaller

The upper and lower limits can be obtained according to the actual situation.

Solve the problem according to the monotonicity of the trigonometric function and determine the definition domain.

(!) Don't mistake people upstairs! Don't mess around if you don't understand! ）

Find the following function zhi to obtain the maximum value of dao and the minimum value of self.

return to the set of variables x, and write the maximum value and minimum value respectively: y=1-1 3*sinx solution: when sinx=-1 y takes the maximum value 4 3, then the set of x is, when sinx=1 y takes the minimum value 2 3, then the set of x is.

2.Monotonic interval: y=-1 2sinxSolution:

y=u 2 is the subtraction function, u=sinx is the increasing function, y=-1 2*sinx is the decreasing function, and its subtraction interval is the increasing interval of sinx, that is, [(2k-1 2) ,2k+1 2) ]k is an integer; In the same way, its increase interval is the subtraction interval of sinx, i.e., [(2k+1 2) ,2k+3 2)].

1. Transform into a trigonometric function.

For example, f(x)=sinx 3cosx=2sin(x 3) has a maximum value of 2 and a minimum value of 2

2. Use the commutation method to become a quadratic function.

For example: f(x)=cosx cos2x

cosx＋2cos²x－1

2t t 1 [where t=cosx [ 1,1]].

Then the maximum value of f(x) is obtained when t=cosx=1, which is 2, and the minimum value is obtained when t=cosx=1 4, which is 9 8

At one time, it can be transformed into a general trigonometric function sin, cos tan to find the maximum and minimum values (range) according to the image

The quadratic one can be changed into a quadratic function by the commutation method, and then the vertex formula can be used to find the maximum and minimum values in the value range, and the commutation can be turned into a checkmark function.

It's all about combining with images.

The maximum value and minimum value of sinx and cosx are all 1, and the trigonometric function is in the form of a·sinx+b or a·cosx+b, the maximum value is a+b, and the minimum value is -a+b

Method 1:

The first step is to define the domain clearly;

The second step is to find out on the diagram.

Method 2: Derivative, which is also the first to find the definition domain.

Then find the extreme point, and you can find the maximum value at the extreme point and the endpoint of the defined domain!