The function f (x) is known to be equal to SiNx plus cosx If f (x) is equal to 2F (- x), find (COS) ² X minus sinxcosx) / 1 plus sin ² Straight of X 2. Find that the function f (x) is equal to f (x) multiplied by F (- x) plus F ² (x) Maximum and monotonically increasing interval of

The function f (x) is known to be equal to SiNx plus cosx If f (x) is equal to 2F (- x), find (COS) ² X minus sinxcosx) / 1 plus sin ² Straight of X 2. Find that the function f (x) is equal to f (x) multiplied by F (- x) plus F ² (x) Maximum and monotonically increasing interval of

1、
f(x)=sinx+cosx
f(x)=2f(-x)
∴sinx+cosx=2[sin(-x)+cos(-x)]
sinx+cosx=﹣2sinx+2cosx
3sinx=cosx
tanx=sinx / cosx=1/3
(cos ² x-sinxcosx)/(1+sin ² x)
=(cos ² x-sinxcosx)/(sin ²+ cos ²+ sin ² x)
=(cos ² x-sinxcosx)/(2sin ²+ cos ²) (fraction divided by COS) ² x. (get)
=(1 - tanx)/(2tan ² x+1)
=6/11
2、
F(x)=f(x)f(-x) + f ² (x)
=(sinx+cosx)(-sinx+cosx) + (sinx+cosx) ²
=cos ² x-sin ² x + sin ² x+cos ² x+2sinxcosx
=2cos ² x+2sinxcosx
=cos2x+sin2x+1
=√2[(√2/2)cos2x+(√2/2)sin2x] + 1
=√2sin(2x + π/4) + 1
The maximum value is √ 2 + 1
-π/2 + 2kπ≤2x + π/4≤π/2 + 2kπ,k∈Z
-3π/4 + 2kπ≤2x≤π/4 + 2kπ,k∈Z
-3π/8 + kπ≤x≤π/8 + kπ,k∈Z
The monotonically increasing interval is [- 3 π / 8 + K π, π / 8 + K π], K ∈ Z

Function f (cosx) = cos3x, then f (SiNx) is equal to

f(cosx)=cos3x
Then f (SiNx)
=f[cos(π/2-x)]
=cos[3(π/2-x)]
=cos(3π/2-3x)
=-sin3x

Function f (x) = cos (2x) / (cosx SiNx), what is the derivative f '(x) equal to?

f(x)=(cos 2x)/(cos x - sin x )=(cos^2 x -sin^2 x )/(cos x -sin x)=cos x + sin x
So f '(x) = - SiN x + cos x

What's the difference between calculus, definite and indefinite integral?

Calculus includes definite integral and indefinite integral. The result of definite integral is a value and the result of indefinite integral is a function

Relationship between definite integral, indefinite integral, calculus

As we all know, the two major parts of calculus are differential and integral. In fact, differential is to find the derivative of a function, and integral is to find the derivative of a known function. Therefore, differential and integral are inverse operations. In fact, integral can also be divided into two parts. The first is simple integral, that is, to find the original function with known derivative. If the derivative of F (x) is f (x), then f (x) + C (C is a constant) The derivative of F (x) is also f (x), that is, integrating f (x) does not necessarily get f (x), because the derivative of F (x) + C is also f (x), and C is an endless constant, so there are countless results of F (x) integration, which is uncertain. We always use f (x) + C instead, which is called indefinite integral
As opposed to indefinite integral, it is definite integral
The so-called definite integral is in the form of ∫ f (x) DX (the upper limit a is written above ∫ and the lower limit B is written below ∫). It is called definite integral because the value obtained after integration is certain, which is a number, not a function. The formal name of definite integral is Riemann integral. See Riemann integral for details. In my own words, It is to divide the image of the function in the rectangular coordinate system into countless rectangles with a straight line parallel to the Y axis and the X axis, and then add up the rectangles in an interval [a, b]. The result is the area of the image of the function in the interval [a, B]. In fact, the upper and lower limits of the definite integral are the two endpoints A and B of the interval

How to find the breakpoint derivative of piecewise function? Must we use the definition method to find the left and right derivatives?

Of course not. As long as the function on an interval can be smoothly extended outside the interval, the unilateral derivative at the end of the interval can be calculated without definition
For example, when x = a, y = g (x) = 2x + 1
In this case, first try to extend F and G near a according to the function expression. It can be found that x = a is the discontinuity of F (x), and the left derivative here should be calculated separately; However, x = a is not the discontinuous point of G (x), and the right derivative can be obtained directly according to the expression
supplement
To xiongxionghy:
Learning and dealing with exams are two different things. Our education system has ruined the form of exams. You should no longer encourage students to only think about dealing with exams when they study. The purpose of learning is to master knowledge, and as long as they really understand it, they will not have unclear ideas and will not be prone to "in case of wrong judgment", Naturally, you will know how to deal with low - level examiners
On this issue, I know that the landlord certainly does not understand the concept of "analytical continuation", so I only give a very rough explanation and bring an example to let him experience it himself