1) Let f (x) = 2arccosx + arctanx - π, and prove that the image of y = f (x) is symmetric about the origin center 2) Find the positive integer x, y satisfying the equation arctanx + arctan (1 / y) = arctan3

1) Let f (x) = 2arccosx + arctanx - π, and prove that the image of y = f (x) is symmetric about the origin center 2) Find the positive integer x, y satisfying the equation arctanx + arctan (1 / y) = arctan3

2. Arctanx + arctan1 / y = arctan3 Tan (arctanx + arctan1 / y) = Tan (arctan3) = 3 (x + 1 / y) / (1-x / y) = 3 Y0 (XY + 1) / (Y-X) = 3 XY + 3x-3y + 1 = 0 (x-3) (y + 3) = - 10

I almost forgot= tan(arcsinx1/3)=------ cos(2sin3/5)=------- cos(1/2arcsin4/5)=-------- sin(2arcsin3/5)=----- cos(arcsin4/5+arcsin8/17)=-------

The solution Tan (arcsin1 / 3) = sin (arcsin1 / 3) / cos (arcsin1 / 3)
＝﹙1／3﹚／[1－﹙1／3﹚]½
＝√2／4
cos﹙2arcsin3／5﹚＝1－ 2sin²﹙arcsin3／5﹚
＝ 1－2×﹙3／5﹚²
＝ 7／25
cos[﹙1／2﹚arsin﹙4／5﹚]＝﹛[1＋cos﹙arcsin4／5﹚]／2﹜½
＝﹛1＋[1－﹙4／5﹚²]½／2﹜½
＝2√5／5
sin﹙2arcsin3／5﹚＝2sin﹙arcsin3／5﹚cos﹙arcsin3／5﹚
＝2×﹙3／5﹚×[1－﹙3／5﹚]½
＝2×﹙3／5﹚×﹙4／5﹚
＝24／25
cos﹙arcsin4／5＋arcsin8／17﹚＝cos﹙arcsin4／5﹚cos﹙arcsin8／17﹚－sin﹙arcsin4／5﹚sin﹙arcsin8／17﹚＝[1－﹙4／5﹚²]½×[1－﹙8／17﹚²]½－﹙4／5﹚×﹙8／17﹚
＝﹙3／5﹚×﹙15／17﹚－﹙4／5﹚×﹙8／17﹚
＝13／85

The range of anti trigonometric function The range of arcsinx, arccosx, arctanx, arccotx The function f (x) = arctanx, when x tends to be positive infinity, what is the value

arcsin:[-pai/2,pai/2]
arccos:[0,pai]
arctan:(-pai/2,pai/2)
artcot:(0,pai)

What is the range of anti trigonometric function?

Inverse trigonometric functions have special marks arcsinx and arctanx, which are strictly defined. Their range of values is of course the main value range. The inverse function of y = SiNx (x ∈ [π / 2,3 π / 2]) cannot be called anti trigonometric function. However, the inverse function of y = SiNx (x ∈ [π / 2,3 π / 2]) cannot be called inverse trigonometric function, Yes can and can only be expressed by the inverse trigonometric function: y = π - arcsinx. - x0d. Therefore, although its value range is not the main value range of the inverse trigonometric function, it is related to the principal value range of the inverse trigonometric function

The determinant is proved to be 0 by linear algebra linear algebra 0 a12 a13 a14 a15; -a12 0 a23 a24 a25; -a13 -a23 0 a34 a35; = 0 -a14 -a24 -a34 0 a45; -a15 -a25 -a35 -a45 0

Therefore, the determinant of D. is unchanged=
0 -a12 -a13 -a14 -a15;
a12 0 -a23 -a24 -a25;
a13 a23 0 -a34 -a35;
a14 a24 a34 0 -a45;
a15 -25 a35 a45 0.
After extracting the common factor-1 in each row, the remaining determinant is the same as the original determinant, so
D=(-1)*(-1)*(-1)*(-1)*(-1)*D=-D,
So d = 0

(linear algebra) using the properties of determinant to prove

After simplifying the determinant property as shown in the figure, there are two columns in proportion, so the determinant is 0. The economic mathematics team will help you solve the problem, please evaluate it in time

Determinant property 5 how to prove, Tongji version of linear algebra

This property is that if the elements of a column (row) are the sum of two numbers, then the determinant can be divided into the sum of two determinants. It can be proved by definition that considering the second definition of determinant (theorem 2), the definition expanded according to the natural order of column index

How is it wrong to prove determinant equality, prove 1 1 1 a b c a³ b³ c³ =（a+b+c)（a-b)(a-c)（c-b)

According to the first column, B-A c-ab-a, C-A, C-A = (B-A) (C-A) - (C-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (B-A) (C-A) (B-A) (C-A) (B-A) (B-A) (C-A) (B-A) B-A C-A (B-A) (C-A) (B-A) (B-A) (B-A) (B-A) (B-A

Proof of determinant The first line ax + ay ay + BZ AZ + BX ay+bz az+bx ax+by az+bx ax+by ay+bz It is proved that he is equal to a + B 3 times a determinant, the first line X Y Z, the second line y z x x, the third line Z x y

Take down the first line first
ax ay az
ay+bz az+bx ax+by
az+bx ax+by ay+bz
+
by bz bx
ay+bz az+bx ax+by
az+bx ax+by ay+bz
And then we split the second row separately
ax ay az
ay az ax
az+bx ax+by ay+bz
+
ax ay az
bz bx by
az+bx ax+by ay+bz
+
by bz bx
ay az ax
az+bx ax+by ay+bz
+
by bz bx
bz bx by
az+bx ax+by ay+bz
And then separate the third row
ax ay az
ay az ax
az ax ay
+
ax ay az
ay az ax
bx by bz
+
ax ay az
bz bx by
az ax ay
+
ax ay az
bz bx by
bx by bz
+
by bz bx
ay az ax
az ax ay
+
by bz bx
ay az ax
bx by bz
+
by bz bx
bz bx by
az ax ay
+
by bz bx
bz bx by
bx by bz
Record D=
x y z
y z x
z x y
In the above eight items:
The first term = a ^ 3 * a,
The second term = 0 (linear correlation between the first and third lines),
The third term = 0 (linear correlation between the second and third lines),
The fourth term = 0 (linear correlation between the first and third lines),
The fifth term = 0 (linear correlation between the first and second lines),
Item 6 = 0 (linear correlation between the first and second lines),
The seventh term = 0 (linear correlation between the second and third lines),
Item 8 = B ^ 3 * a (swap the first and second rows and then the new first and third rows)
Therefore, the value of the determinant = (a ^ 3 + B ^ 3) * a, that is to say
(in your result, a should be changed to a ^ 3. A, B is symmetrical, so it can't be a + B ^ 3)

Prove the following determinant The main diagonal is cosx, 2cosx, 2cosx, 2cosx. The diagonal lines on both sides are 1. The others are all 0. It is proved that he is equal to cosnx

By mathematical induction, when n = 2, it holds. The following hypothesis holds for n ≥ 2, considering the case that N + 1 is, there is
That is, when n + 1, the conclusion also holds, so the original conclusion holds