The unit of the graph is shifted to the left by the unit of the graph

The unit of the graph is shifted to the left by the unit of the graph

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0

The image of the function y = sin2x shifts π to the left
Four units give y = sin (2x + π)
2) And then move up one unit to get y = sin (2x + π)
2）+1=1+cos2x=2cos2x
So the answer is: y = 2cos2x

Given the graph of the first order function passing through a (2, - 1) and B, where point B is the intersection point of another straight line y = - 1 / 2x + 3 and the Y axis, find the expression of this function

Because point B is the intersection of another straight line y = - 1 / 2x + 3 and the Y axis
So B coordinates (0, 3)
Let the first degree function expression y = KX + 3
Because of a (2, - 1)
So - 1 = 2K + 3
k=-2
So the first-order function expression y = - 2x + 3

What equation is the square of X + 1 of the square of X

If it is written as x ^ 2 + 1 / x ^ 2 = 0, note: I added "= 0", this is a fractional equation, because if there is no "=", it is not an equation at all. If it is written as y = x ^ 2 + 1 / x ^ 2, note: I added "y =" before your formula, because if there is no "y =", it is not a function at all. It has no specific name, because in

Find the maximum and minimum of the function y = 3 - 5 / (2 + SiNx)~

Because - 1 ≤ SiNx ≤ 1, so: 1 ≤ 2 + SiNx ≤ 3, so 1 / 3 ≤ 1 / (2 + SiNx) ≤ 1, so: - 5 ≤ - 5 / (2 + SiNx) ≤ - 5 / 3, so: - 2 ≤ 3 - 5 / (2 + SiNx) ≤ 4 / 3, that is: the maximum value is 4 / 3, the minimum value is - 2

t->0,lim[tan(sinx)-sin(tanx)]/(tanx-sinx)=?

The original formula = Lim {X - > 0} {Tan (SiNx) - Tan (TaNx) [1 + cos (TaNx) - 1]} / (TaNx SiNx)
=lim{x->0}{tan(sinx-tanx)[1+tan(sinx)tan(tanx)]/(tanx-sinx)
-lim{x->0}tan(tanx)[cos(tanx)-1]/(tanx-sinx)
Because Tan (SiNx TaNx) ~ SiNx TaNx and Tan (TaNx) ~ TaNx (x - > 0), the above formula
=-lim{x->0}[1+tan(sinx)tan(tanx)]-lim{x->0}[cos(tanx)-1]/(1-cosx)
Because cos (TaNx) - 1 ~ - (TaNx) ^ 2 / 2 (x - > 0) and 1-cosx ~ - x ^ 2 / 2 (x - > 0), the above formula
=-1+lim{x->0}(tanx/x)^2
=-1+1
=0

0

∵π/2

In the acute triangle ABC, edges a and B are two of the equations x ^ 2-2, the root sign 3x + 2 = 0, and the angles a and B satisfy 2Sin (a + b) - radical 3 = 0: then the area of the triangle is

Solution in △ ABC, angle c = 180 - (a + b)
Sinc = sin (a + b), since 2Sin (a + b) - radical 3 = 0, sin (a + b) = root 3 / 2
In other words, sinc = root 3 / 2, because the angle c is an acute angle
Therefore, the angle c = 60 degrees
Because, a and B are two of the equations x ^ 2 - (2 radical 3) x + 2 = 0
Therefore, a * b = 2
Moreover, the angle c is the angle between side a and side B
Because, the area of △ ABC s = (1 / 2) * a * b * sinc (sinc = sin60 = 1 / 2)
That is, s = (1 / 2) * 2 * radical 3 / 2
Therefore, s = root 3 / 2 (area unit)
Answer: angle c = 60 degrees
Area of triangle s = root 3 / 2 (area unit)

In △ ABC, BC = a, AC = B, a, B are two roots of x ^ 2-2 times the root sign 3x + 2 = 0, and 2cos (a + b) = 1. Find (1) is the degree of C (2) The length of AB (3) △ the area of ABC

1. Cos (a + b) = 1 / 2, cos (π - C) = 1 / 2
cosc=-1/2
So ∠ C = 120 degrees
2. From the equation, we can get two solutions which are √ 3 + 1 or √ 3 - 1
Cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2abcosc
2 + 2
c=√10
3、S△ABC=0.5absinc=0.5×2×√3/2=√3/2

[2sin50 degree + sin10 degree (1 + √ 3tan10 degree) √ (1 + cos20 degree) Is root sign

[2sin50 degrees + sin10 degrees (1 + √ 3tan10 degrees)] √ (1 + cos20 degrees) = [2sin50 degrees + sin10 degrees (1 + √ 3tan10 degrees)] √ (1 + cos20 degrees) = [2sin50 degrees + sin10 degrees (cos10 + √ 3sin10) / cos10] √ (1 + cos20 degrees) = [2sin50 degrees + sin10 degrees * 2 * (1 / 2cos10 + √ 3 / 2 *)