It is proved that (SiNx + siny) / cosx cosy = CTG (Y-X) / 2

It is proved that (SiNx + siny) / cosx cosy = CTG (Y-X) / 2

（sinx+siny)/(cosx-cosy)
=[2sin(x+y)/2cos(x-y)/2]/[-2sin(x+y)/2sin(x-y)/2]
=-[cos(x-y)/2]/[sin(x-y)/2]
=-cot(x-y)/2
=cot(y-x)/2

Using Lagrange mean value theorem to prove inequality |sinx-siny| ≤ |x-y|

Let f (x) = SiNx
Then f '(x) = cosx
There is ξ between X and y,
bring
sinx-siny=f '(ξ)(x-y)
=cosξ(x-y)
So,
|sinx-siny|=|cosξ(x-y)|
≤|x-y|

If the image of function y = sin (x + π / 6) is translated according to vector a = (- π, 0), then the image of function after translation is transformed A. On (- π / 6,0) symmetry B. On the symmetry of the line x = π / 6 C. On point (π / 3,0) symmetry D. On the symmetry of the line x = π / 2 Seek the process!

Solution: the maximum value and minimum value of string function are axisymmetric, and when the function value is 0, the center is symmetric
The direction of the known vector is moving along the negative half axis of the X axis, which is reflected in the function, and the independent variable plus π
The image with y = sin (x + π / 6) is translated by vector a = (- π, 0) and y = sin (x + π 7 / 6) = - sin (x + π / 6)
A term, x = - π / 6, y = 0
B, coordinate substitution can not get the maximum value, wrong
Item C, put in, wrong
D term, x = π / 2 in, wrong

Let's take a function image by vector a=（π After 3, - 2) translation, the expression of the image is y = sin (x + π) 6) - 2, then the analytic formula of the original function is______ .

According to the meaning of the title, y = g (x) = sin (x + π)
6) - 2 by vector
b=（-π
3,2) translation, the analytic expression of the original function can be obtained,
The original function y = sin [(x + π)
6）-（-π
3）]-2+2
=sin（x+π
2）
=cosx．
So the answer is: y = cosx

Trigonometric function translation problem! All the points in the image of sin2x are moved to the left parallel by π / 6 unit length, and then the abscissa of each point in the image is extended to 2 times of the original. The analytic formula of the image is? A.y=sin(x-π/3） B.y=sin(x+π/3) C.y=sin(2x+π/6） D.y=sin(4x+π/3)

Left plus right minus
So after the left shift is sin2 (x + π / 6)
The abscissa is twice as big, so it's sin2 (x / 2 + π / 6)
Choose B

The monotone decreasing interval of 2x-3x + 1 under the root of the culvert number f (x) = is

F (x) = 2x-3x + 1 = under radical [2 * (x ^ 2-3x / 2 + 9 / 16-9 / 16) + 1]
=[2] / 2
To monotonically decrease: x = 0
The solution is: X

F (x) = the monotone decreasing interval of (2x ^ 2-3x-2) under the radical sign is?

First of all, a monotone function is defined. After the root sign is added, it is still a monotone function with invariable monotonicity in the definition domain. But the key to solve this kind of problem is to pay attention to the definition domain

Monotonicity of function f (x) = 1-2x under radical_____ The (increase or decrease) range is_____

F (x) = 1-2x under the radical sign, whose definition domain is (- ∞, 1 / 2],
Obviously, 1-2x is monotonically decreasing on (- ∞, 1 / 2],
So 1-2x under the radical is monotonically decreasing on (- ∞, 1 / 2],
So the monotone decreasing interval of F (x) is (- ∞, 1 / 2]

Write the monotone interval (1) f (x) = x 2 - 2x (2) f (x) = / x? - 2x/

(1) F (x) = √ (x ^ 2-2x) is a composite function, and the outer layer y = √ u is an increasing function

Given the function f (x) = 2 (x power) + root sign (1-x Square) g (x) = 3x + 1-radical (1 - (the square of x)), find the maximum value of F (x) + G (x) F (x) = 2x ^ 2 + (1-x ^ 2) ^ 1 / 2 g (x) = 3x + 1 - (1-x ^ 2) ^ 1 / 2

F (x) = 2 (x power) + root sign (1-x Square) g (x) = 3x + 1-radical (1 - (the square of x)). First, find the definition field: 1-x? ≥ 0, and get - 1 ≤ x ≤ 1. The domain is [- 1,1]; f (x) + G (x) = 2 ^ x + √ (1-x?) + 3x + 1 - √ (1-x?) f (x) + G (x) = 2 ^ x + 3x + 1, (x ∈ [- 1,1])