Our ship is 50 degrees south by west of the enemy island a, 12 kilometers away from B. We find that the enemy ship is sailing from the island along the direction of 10 degrees north by west at a speed of 10 kilometers per hour. What is the speed required for our ship to catch up with the enemy ship in 2 hours? Ship a finds that ship B is 60 degrees north by east of ship a, which is a nautical mile away from ship A. ship B is moving northward. The speed of ship a is three times that of ship B's root sign. What's the fastest way for ship a to catch up with ship B? How far is it from ship a to catch up with ship B?

Our ship is 50 degrees south by west of the enemy island a, 12 kilometers away from B. We find that the enemy ship is sailing from the island along the direction of 10 degrees north by west at a speed of 10 kilometers per hour. What is the speed required for our ship to catch up with the enemy ship in 2 hours? Ship a finds that ship B is 60 degrees north by east of ship a, which is a nautical mile away from ship A. ship B is moving northward. The speed of ship a is three times that of ship B's root sign. What's the fastest way for ship a to catch up with ship B? How far is it from ship a to catch up with ship B?

The first question, if we catch up at point C, then the enemy ship has traveled 10 * 2 = 20 km, and the distance between a and our ship (set as point B) is 12, then we form a triangle, that is, the angle cab is 120 degrees, 180-10-50 = 120 degrees, AC = 20, ab = 12. In this way, we should be able to calculate the distance of BC (mainly because I forgot the formula for a long time after graduation), and then / 2 should be able to get it
The second problem is almost similar to the solution, the formula I forget is not good, how to say oh

tan²α-sin²α=tan²α×sin²α
（cosα-1）²+sin²α=2-2cosα
Sin quartic x + cos quartic x = 1-2sin & # 179; xcos & # 178; X
The solution of known Tan α - 3 is: ① 4sin α - 2cos α / 5cos α + 3tan α; ② sin α cos α; ③ (sin α + cos α) &;
Given cos α = 1 / 4, find sin α and Tan α
Judge the sign of the following trigonometric functions

(1) Left = (Sina) ^ 2 / (COSA) ^ 2 - (Sina) ^ 2 = (Sina) ^ 2 * [1 / (COSA) ^ 2-1] = = (Sina) ^ 2 * [(1 - (COSA) ^ 2) / (COSA) ^ 2] = right (2) left = (COS β) ^ 2-2cos β + 1 + (sin β) ^ 2 = right (3) left = [(SiNx) ^ 2 + (cosx) ^ 2] ^ 2-2 (SiNx) ^ 2 * (cosx) ^ 2 = right (4) should be

(1) 3 / 2 times cosx - √ 3 / 2 times SiNx;
(2) 3 times SiNx / 2 + cosx / 2;
(3) 2 / 4 times sin (π / 4 - x) + 6 / 4 times cos (π / 4 - x);

(1) 3 / 2 times cosx - √ 3 / 2 times SiNx
=√ 3 (√ 3 / 2 times cosx - 1 / 2 times SiNx)
=3 (cos30 times cosx - sin30 times SiNx)
=√3cos（30+x）
(2) √ 3 times SiNx / 2 + cosx / 2 = 2 (√ 3 / 2 times SiNx / 2 + 1 / 2cosx / 2) = 2Sin (x / 2 + 30)
(3) √ 2 / 4 times sin (π / 4 - x) + √ 6 / 4 times cos (π / 4 - x)
=√2/2（cos60sin（π/4 - x）+sin60cos（π/4 - x））
=√2/2sin（7π/12-x）

On the periodicity and symmetry of functions
1. The function f (x) satisfies f (0.5 + x) = f (0.5-x) for all real numbers x, and the equation f (x) = 0 has three real roots whose sum is
2. The real roots of the equations x ^ 5 + X + 1 = 0 and X + x ^ 0.2 + 1 = 0 are a, B, a + B, respectively=
3. If the images of the functions y = f (x), y = f (- x), y = - (- x) defined on R coincide, then the range of the function y = f (x) is
4. The function y = f (x) is an even function, and its period is 2. When x belongs to [2,3], there are two points a and B on the image of F (x) = X-1, y = f (x). Their ordinates are equal (point a is on the left side of point B), their abscissa are on the interval [1,3], and the coordinates of fixed point C are (0, a), where a > 2. Find the maximum area of triangle ABC
(1, 2, 3 can not process, 4 must write in full,

1.f(0.5+x)=f(0.5-x)
It is concluded that f (x) = f (1-x)
So the sum of the three real roots is 1 + 0.5 = 1.5
2. X ^ 5 + X + 1 is monotone, and a and B ^ 0.2 are his roots, with a = B ^ 0.2
So we get a + B = - 1
3. Y = f (x), y = f (- x) images coincide, indicating that f (x) is symmetrical about y axis;
Y = f (- x), y = - f (- x) images coincide, which shows that f (x) is symmetrical about X axis, so f (x) is equal to 0, so the value range is {0}
4. X belongs to [0,1], f (x) = x + 1; X belongs to [- 1,0], f (x) = - x + 1
So x belongs to [1,2], f (x) = - x + 3
Let the ordinates of a and B be t, then s = 1 / 2 (2t-2) (A-T)

Proof of function symmetry
If the function y = f (x) satisfies: F (a + x) = f (b-X), then whose symmetry is the image of the function

For any x, let x = X1 - (B-A) / 2, then x 1 = x + (B-A) / 2, f (a + x1) = f (b-x1), that is, f (a + X + (B-A) / 2) = f (b-X - (B-A) / 2), that is, f ((a + b) / 2 + x) = f ((a + b) / 2-x)

Given that f (x) is an odd function on R, and if x ∈ (- ∞, 0), f (x) = - xlg (2-x), find the analytic expression of F (x)

The solution ∵ f (x) is an odd function, and we can get f (0) = - f (0), f (0) = 0. When x ∵ 0, - x ∵ 0, f (- x) = xlg (2 + x), f (x) = xlg (2 + x) (x ∵ 0).. f (x) = − xlg (2 − x) & nbsp; & nbsp; & nbsp; & nbsp; (x ＜ 0) − xlg (2 + x) & nbsp; & nbsp; & nbsp; (x ≥ 0), that is, f (x) = - xlg (2 + | x ∈ R)

Let f (x) and G (x) be defined by X ∈ R and X ≠± 1, f (x) be even, G (x) be odd, and f (x) + G (x) = 1 x − 1. Find the analytic expressions of F (x) and G (x)

∵ f (x) is an even function, G (x) is an odd function, ∵ f (- x) = f (x), and G (- x) = - G (x) by F (x) + G (x) = 1x − 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;, we get f (− x) + G (− x) = 1 − x − 1, that is, f (x) − g (x)

The even function f (x) defined on [- 2,2] decreases monotonically on the interval [0,2]. If f (1-m) < f (m), then the value range of real number m is ()
A. m＜12B. m＞12C. -1≤m＜12D. 12＜m≤2

∵ function is even function, ∵ f (1-m) = f (| 1-m |), f (m) = f (| m |), ∵ even function f (x) defined on [- 2,2] monotonically decreases on interval [0,2], f (1-m) < f (m), ∵ 0 ≤| m | < | 1-m | ≤ 2, then - 1 ≤ m < 12

Senior high school mathematics -- function parity
Let y = f (x) [x belongs to R, and X is not equal to 0] for any nonzero real number x, y, f (XY) = f (x) + F (y) holds
1. Prove that f (- 1) = f (1) = 0, and f (1 / x) = - f (x) [x is not equal to 0]
2. Judge the parity of function
3. If f (x) increases monotonically from 0 to positive infinity, the solution to the inequality f (1 / x) - f (2x-1) is not less than 0

1. If f (1 * 1) = f (1) + F (1), then f (1) = f (1) + F (1), so f (1) = 0f (- 1 * 1) = f (- 1) + F (1), then f (- 1) = f (- 1) + F (1) desired f (- 1) = 0 when x is not equal to 0; f (1) = f (1 / X * x) = f (1 / x) + F (x) = 0, so f (1 / x) = - f (x) [x is not equal to 0] 2, because f (- x) = f (- 1 * x) = f (- 1) + F

F (x) is an odd function and G (x) is an even function, satisfying f (x) + G (x) = (a ^ x) - A ^ (- x) + 2
If G (b) = a, then f (2) =?

4 - (1 / 4) if f (x) is an odd function and G (x) is an even function, then f (- x) = - f (x), G (- x) = g (x) replace X in F (x) + G (x) = (a ^ x) - A ^ (- x) + 2 with - x, then f (- x) + G (- x) = [a ^ (- x)] - A ^ (x) + 2, that is - f (x) + G (x) = [(a ^ (- x)]) - A ^ (x) + 2