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The range of y = Sin & # 178; X + 2sinxcosx + 1
Y = Sin & # 178; X + 2sinxcosx + 1 = [1-cos (2x)] / 2 + sin (2x) + 1 = sin (2x) - (1 / 2) cos (2x) + 3 / 2 = √ [1 & # 178; + (- 1 / 2) &# 178;] sin (2x - θ) + 3 / 2, where Tan θ = 1 / 2 = (√ 5 / 2) sin (2x - θ) + 3 / 2-1 ≤ sin (2x - θ) ≤ 1 (3 - √ 5) / 2 ≤ (√ 5 / 2) sin (2x - θ
Find the range y = cos & # 178; X + 2A sin (x) + 1
y=cos²x+2asinx+1
=1-sin²x+2asinx+1
=-(sinx-a)²+2+a²
When a > 0, the maximum value of Y is y = 2 + A & # 178; - (1-A) &# 178; = 1 + 2A
The minimum value of Y is y = 2 + A & # 178; - (- 1-A) &# 178; = 1-2a
When a
The minimum positive period sum range of y = sin ^ 2 (x + π / 2) - Sin ^ 2 (x - π / 4)
y=sin²(x+π/2) - sin²(x-π/4)=cos²x + [1-2sin²(x-π/4)-1]/2=(2cos²x-1+1)/2 + [cos2(x-π/4)-1]/2=1/2cos2x+1/2 + 1/2cos(2x-π/2)-1/2=1/2cos2x + 1/2sin2x=√2/2(√2/2cos2x+√2/2si...
The range of function y = sin2 (π 4 + x) - sin2 (π 4-x) is ()
A. [-1，0]B. [0，1]C. [-1，1]D. [−12，1]
∵ y = sin2 (π 4 + x) - sin2 (π 4-x) = 1 − cos (π 2 + 2x) 2-1 − cos (π 2 − 2x) 2 = 1 + sin2x2-1 − sin2x2 = sin2x, the value range of the function y = sin2 (π 4 + x) - sin2 (π 4-x) is [- 1, 1]
In the following functions, the minimum positive period is π, and the tangent image is symmetric with respect to the line x = π / 3?
A y=sin(2x-3)
B y=sin(2x-π/6)
C y=sin(2x+π/6)
D y=sin(x/2+π/6)
Can you write down the steps? Thank you
In the following functions, the function whose period is 4 π and the image is symmetric with respect to the line x = π / 3 ()
A.y=2sin(x/2-π/3)
B.y=2sin(x/2-π/6)
C. Y = 2Sin (2x - π / 3) write steps, thank you
D.y=2sin(x/2-π/3)
sorry. Correct: a.y = 2Sin (x / 2 + π / 3)
B.y=2sin(x/2-π/6)
C.y=2sin(2x+π/3)
D. Y = 2Sin (x / 2 - π / 3) gives D, y = - 1
Obviously, the periods of a, B and D are 2 π / (1 / 2) = 4 π,
To make the image symmetrical about the straight line x = π / 3, substitute x = π / 3 into the analytical formula, y can take the maximum or minimum value, and then substitute it into the verification, but your analytical formula A and D are the same, please modify the title and do it yourself
In the following functions, where the minimum positive period is π and the image is axisymmetric with respect to the line x = π 3 ()
A. y=sin（2x-π3）B. y=sin（2x+π6）C. y=sin（2x-π6）D. y=sin（12x+π6）
For y = sin (2x - π 3), its period is π. When x = π 3, the function y = sin (2x - π 3) = 32 is not the maximum value, so the image of the function is not axisymmetric with respect to the line x = π 3, so a is excluded. For y = sin (2x + π 6), its period is π. When x = π 3, the function y = sin (2x - π 3) = 12 is not the maximum value, so the image of the function is not axisymmetric with respect to the line x = π 3, so a is excluded Except B. because the period of function y = SinSin (2x - π 6) is 2 π 2 = π, when x = π 3, the function y = sin (2x - π 6) = 2 gets the maximum, so the image of function y = sin (2x - π 6) is axisymmetric with respect to the line x = π 3, so C satisfies the condition. For y = sin (12x + π 6), because the period of function is 2 π 12 = 4 π, it does not satisfy the condition, so D is excluded
The function image of sin, period π, maximum y is 3, symmetry axis is π / 6?
As the title, I mainly forget how to find the axis of symmetry
Let it be asin (Wx + b)
w=2π/T=2π/π=2
A=3
The axis of symmetry of SiNx is x = k π + π / 2; K belongs to Z
2x+b=kπ+π/2
x=(kπ+π/2-b)/2
Then sin (2x + b) symmetry axis is x = (K π + π / 2-B) / 2
x=(kπ+π/2-b)/2=π/6
b=kπ+π/2-π/3
=kπ+π/6
Let k = 0, B = π / 6
The original function is 3sin (2x + π / 6)
In mathematics, two functions are used to judge the formula of the symmetry of a function with respect to a straight line, as well as the formula for judging the period,
For example, y = f (A-X) and y = f (B + x) are symmetric with respect to the line x = (a + b) / 2,
In F (A-X) = f (B + x), X is replaced by X-B, and f (a + b-X) = f (x) is obtained. Let (m, n) be any point in the image of y = f (x), then n = f (m) can be easily obtained. The symmetric point of (m, n) with respect to the straight line x = (a + b) / 2 is (a + B-M, n) and N = f (m) = f (a + B-M) so that the point (a + B-M, n) is also on the image of y = f (x), so the image of F (x) is symmetric with respect to x = (a + b) / 2
Periodic formula of function
Why f (x + a) = - f (x) period is 2A
Because f (x + a) = - f (x)
And f (x) = - f (x-a)
So f (x + a) = f (x-a)
That is, f (x + 2a) = f (x)
So the period is 2A
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