Solve a simple trigonometric function value problem

Let's start by finding a that makes the function f(a) minimal.

The square root is a monotonic function, so making f(a)=sqrt( cos(a) 2 + cos(2 3* -a) 2 ) minimum a is the same as making the function g(a) = cos(a) 2 + cos(2 3* -a) 2 minimum a. So, we don't need to think about the root number.

Based on cos(2x)=2cos(x) 2-1, we know that cos(x) 2 = (1 2)(1+cos(2x)). So, g(a) = cos(a)2 + cos(2 3* -a)2

1/2+(1/2)cos(2a)+1/2+(1/2)cos(4π/3-2a)

1+(1/2)[cos(2a)+cos(4π/3-2a)]

1+(1/2)[cos(2a)+cos(4π/3)cos(2a)+sin(4π/3)sin(2a)]

1+(1/2)[cos(2a)-(1/2)cos(2a)-(sqrt(3)/2)sin(2a)]

1+(1/2)[(1/2)cos(2a)-(sqrt(3)/2)sin(2a)]

1+(1/2)[cos(π/3)cos(2a)-sin(π/3)sin(2a)]

1+(1/2)cos(π/3+2a).

Since the cos(x) function obtains the minimum value at x=, g(a) and f(a) obtain the minimum value at 3+2a=. The solution satisfies a (0,2 3*) and has a value of 3. Substituting this value into f(a) gives you the minimum value (I'm sure you've already figured it out).

f(a)=sqrt( cos(a)^2 + cos(2/3*π-a)^2 )

2cos [(a)^2+(2π/3-a)^2]/2cos [(a)^2-(2π/3-a)^2]/2

There is a graph to know the minimum value of the function f(a) when a=1 3.

f(a)=√[2cos [(/3)^2+(2π/3-π/3)^2]/2cos [(/3)^2-(2π/3-π/3)^2]/2

2cos (π/3)^2cos 0]=√[2cos 3600]=√2

Landlord, how do you know to take the minimum value on 3? You can't just rely on this picture.

There are also 1L, 2L and two.

1l, penultimate step, cos ( 3) 2=1?

2l，2cos(1/3π)^2=4cos(1/3)？

The first type of question is the auxiliary angle formula model.

This is the most common type, and the focus of this model is to be able to convert two times into one, mainly using the double angle formula, and the half-angle formula. It then becomes the structure of the auxiliary angle formula, as in Example 1. Draw an image easily, solve the maximum value problem, and sometimes pay attention to the range of values of the independent variables.

The second type of question is a one-dimensional quadratic function model.

That is to say, no matter how the function is simplified, it cannot be turned into the structure of the auxiliary angle, so some students wonder if they have miscalculated, and it is still like this after a few more calculations, and then Tong Ji Hail wonders if the problem is wrong, don't be funny, the probability of making a mistake is lower than winning the lottery. Take a closer look, when the same order is contradicted by the same angle, see if it is the structure of a quadratic equation when it is different, and if it is, you can also use the image of the quadratic function to find the maximum value.

The third question type, which is less common, is the fractional type, where the numerator has a square term and the denominator is in the form of a product. Take a closer look at the process of solving this example problem. The final solution turned out to be the mean theorem.

The fourth question type is the overall change type of the bureau sail, which is not easy to think of, and it is particularly simple after the change.

The fifth type, the method of selecting the principal element, takes the function as a whole as a parameter, and since this parameter is subject to other constraints, it is found that its range is the result of the requirement.

sin(a) cos = 1 2sin2a according to: sin ·cos = (1 2) [sin( +sin( - available: sina cosa = (1 2) [sin(a+a) + sin(a-a)].

1/2sin2a

and Angle Formula:sin ( sinα ·cosβ ±cosα ·sinβsin ( sinα ·cosβ ·cosγ +cosα ·sinβ ·cosγ +cosα ·cosβ ·sinγ -sinα ·sinβ ·sinγ

cos ( cos sin sin tan ( tan tan ) 1 or dry tan tan )

y=cos^3

x+sin^2

x-cosx=cosx(cos^2x-1)+sin^2x=(1-cosx)sin^2x=(1-cosx)^2(2+2cosx)/2

4=1-cosx+1-cosx+2+2cosx>=3 cubic root [(1-cosx) 2(2+2cosx)].

So the cube root (1-cosx) 2(2+2cosx)<=4 (when 1-cosx=2+2cosx

cosx=-1 3).

So when cosx=-1 3 epoch in, the maximum is 32 27

y=2cosx-3sinx

13(2/√13cosx-3/√13sinx)=√13sin(t-x)

where sint=2 13, cost=3 13

Therefore, when t-x= 2, y=2cosx-3sinx is maximized.

So tanx=tan(t- 2).

cott=-cost/sint

3/2,

Derivative-2sinx-3cosx=0

Sometimes there is the best value. Get tanx=-3 2

Let's talk about whether the maximum value is the largest or the smallest.

tanx=-3 2.

x in quadrant 24.

It is found that tanx=-3 2 always holds at the minimum maximum.

Trigonometric functions are a class of functions in mathematics that belong to the transcendental functions of elementary functions. Their essence is a mapping between a set of arbitrary angles and a set of variables with a ratio. The usual trigonometric function is defined in a planar Cartesian coordinate system, which defines the entire field of real numbers.

Another definition is in a right triangle, but not completely. Modern mathematics describes them as the limits of an infinite series of numbers and the solution of differential equations, extending their definition to complex systems.

Due to the periodic nature of trigonometric functions, it does not have an inverse function in the sense of a single-valued function.

Trigonometric functions have important applications in complex numbers. In physics, trigonometric functions are also commonly used tools.

It has six basic functions:

The name of the function. Sine.

Cosine. Tangent.

Cotangent. Secant.

Cosecant. Symbol.

sincos

tancot

seccsc

Sine function. sin（a）=a/h

Cosine function. cos（a）=b/h

Tangent function. tan（a）=a/b

Cotangent function. cot（a）=b/a

Attached: Some special trigonometric values.

sin0=0

cos0=1

tan0=0

sin15 = (root number 6 - root number 2) 4

cos15 = (root number 6 + root number 2) 4

tan15 = sin15 cos15 (do the math yourself) sin30 = 1 2

cos30 = root number 3 2

tan30 = root number 3 3

sin45 = root number 2 2

cos45=sin45

tan45=1

sin60=cos30

cos60=sin30

tan60 = root number 3

sin75=cos15

cos75=sin15

tan75 = sin75 cos75 (compare yourself) sin90 = cos0

cos90=sin0

tan90 is meaningless.

sin105=cos15

cos105=-sin15

tan105=-cot15

sin120=cos30

cos120=-sin30

tan120=-tan60

sin135=sin45

cos135=-cos45

tan135=-tan45

sin150=sin30

cos150=-cos30

tan150=-tan30

sin165=sin15

cos165=-cos15

tan165=-tan15

sin180=sin0

cos180=-cos0

tan180=tan0

sin195=-sin15

cos195=-cos15

tan195=tan15

sin360=sin0

cos360=cos0

tan360=tan0

PS: In fact, as long as you memorize 0, 30, 45, 60 is enough, and the rest can be calculated by the induction formula.

As shown in the figure below, f(x) is first transformed, then the two corners are used to sum the formula into a single trigonometric formula, and then the domain is evaluated according to the defined domain:

This is a formula in the form of y=asin(wx+a) using the double angle formula and the auxiliary angle formula, and combining the image and properties of the trigonometric function.

First, f(x) is reduced to a trigonometric function of an angle, and then the composite function is used to find its value range.

Calculations such as the above floors.

The answer is as follows: f(x).

cos⁴x-2sinxcosx-sin⁴xcos⁴x-sin⁴x-2sinxcosx(cos²x+sin²)(cos²x-sin²x)-2sinxcosx

cos²x-sin²x)-sin2x

cos2x-sin2x

2[(√2/2)cos2x-(√2/2)sin2x]√2cos(2x+π/4)

So the f(x) maximum is 2.