If the value of the function y = (a + b) cosx ^ 2 + (a-b) SiNx ^ 2 (x belongs to R) is equal to 2, then the coordinates of the symmetric point of point P (a, b) about the origin are?

If the value of the function y = (a + b) cosx ^ 2 + (a-b) SiNx ^ 2 (x belongs to R) is equal to 2, then the coordinates of the symmetric point of point P (a, b) about the origin are?

y=(a+b)cos²x+(a+b)sin²x-2bsin²x
=(a+b)[cos²x+sin²x]-2bsin²x
=(a+b)-2bsin²x=2
So a + B = 2
-2b=0
b=0,a=2
So the coordinates of the symmetric point about the origin are (- 2,0)

The mathematical equation cosx / (1-sinx) = 2 / (π - 2) is solved and X is obtained

Let Tan (x / 2) = t and t ≠ 1
So cosx = (1-T ^ 2) / (1 + T ^ 2)
sinx=(2t)/(1+t^2)
cosx/(1-sinx)=(1-t^2)/(1+t^2-2t)=(1+t)/(1-t)=2/(π-2)
So t = (4 - π) / π = 4 / π - 1
x=2arctan(4/π-1)

Integral of Mathematics (SiNx) ^ 2 - (cosx) ^ 2 / (SiNx) ^ 4 + (cosx) ^ 4

[(sinx)^2-(cosx)^2]/[sinx^4+cosx^4]=-cos2x/[(sinx^2+cosx^2)^2-2sinx^2cosx^2]=-cos2x/1-(sin2x)^2/2∫[(sinx)^2-(cosx)^2]dx/[sinx^4+cosx^4]=∫-cos2xdx/[1-(sin2x)^2/2]=-∫dsin2x/(2-(sin2x)^2)=[-1/(2√2)]...

Finding the solution of the equation SiNx + cosx + 1 = 0

Analysis: the square of SiNx + cosx + 1 = 0, the square of SiNx > = 0 (when x = 0 degree, sin = 0), the square of SiNx + 1 > = 1, and - 1=

The function of image symmetry about origin must be odd function
That's right

This is necessary. The image of odd function must be symmetric about the center of origin (0,0)

Why is f (x) symmetric about the origin if it is an odd function

The judgment of odd function is f (x) = - f (- x). From the above, we can see that the values of F (x) and f (- x) are opposite to each other. Here, X and - X are opposite to each other, right
According to the formula, if you take (x, y) and (- x, - y), you can see that the two points are symmetrical about the origin at a glance. (in the above, we just take X and - x randomly to ensure that they can represent any pair of opposite numbers in the domain of definition, and the result is not for a particular value, and the theory is still valid.)
It should be mentioned that the points constitute lines. Since every pair of points is symmetrical about the origin, the image is also symmetrical about the origin. Even if the image is an infinite number of isolated points without lines, it is still satisfied
Of course, this is just a common case. If you want to be non confrontational, for example, if the definition domain 0 is (- ∞, - x) and (0, x), you can use the y = x image to test. It is not difficult to see that there is no image corresponding to the relative region, or the corresponding is incomplete, but this kind of image is only a special case

If the image of the function y = cos (2x + π / 4) is shifted to the left by π / 4 unit length, the curve C1 is obtained, and C1 and C2 are symmetric about the origin, then the corresponding analytical expression of C2 is

C1 is y = cos (2 (x + π / 4) + π / 4) = cos (2x + 3 π / 4)
So C2 is - y = cos (2 (- x) + 3 π / 4), that is y = - cos (2x-3 π / 4)

If the image of the function y = 3 / (X-2) is shifted to the left by M (M > 0) units, the image is symmetric about the origin or center, then the real number M=_______
It's a process!

The image of the function y = 3 / (X-2) is shifted m (M > 0) units to the left to get y = 3 / (x + m-2)
Since the image is symmetrical about the origin, m-2 = 0, that is, M = 2

After the image of the function y = (3 + a · 3 ^ x) / (3a-3 ^ x) is shifted one unit to the left, the image C1 of y = f (x) is obtained. If the curve C1 is symmetric about the origin, then a
The answer is 1 or - 1, but I think that when a = 1, the denominator cannot be 0, so x ≠ 1, but x can be equal to - 1, then C1 cannot be said to be symmetric about the origin

Pro, the symmetry about the origin is C1, not to y = (3 + a · 3 ^ x) / (3a-3 ^ x)
The translated function f (x) = [3 + A * 3 ^ (x + 1)] / [3a-3 ^ (x + 1)]
When a = 1, x = 0 is the point where the denominator is 0

Y = 1 / 2cos ^ x + radical 3 / 2sinxcosx + 1. X belongs to R. how can the function image be translated and extended from the image of y = SiNx

You're missing a cos square|||
From the title meaning y = 1 / 4cos2x + radical 3 / 4sin2x + 5 / 4
=1/2sin(2x+pi/6)+5/4
By the law of expansion and contraction
The function image can be obtained by y = SiNx moving pi / 6 units to the left, reducing the abscissa by 1 / 2, reducing the ordinate by 1 / 2, and moving up by 5 / 4 units