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If the image of function y = sin (2x + I) shifts π / 6 units to the left and coincides with the image of y = sin2x, then the minimum positive value of I is
A:
The image of y = sin (2x + L) shifts to the left π / 6 units and then coincides with y = sin2x
Therefore, when x = π / 6, y = sin (2x + L) = sin (π / 3 + L) = 0
So: π / 3 + L = π
So: l = 2 π / 3
So: the minimum positive value of L is 2 π / 3
The image of the function y = sin (2x + θ) shifts π to the left The minimum positive value of θ is () A. π Three B. 5π Six C. 4π Three D. 5π Three
The image of ∵ function y = sin (2x + θ) shifts π to the left
Six
∴y=sin2x=sin（2（x+π
6）+θ）=sin（2x+θ+π
3）
That is, 0 + 2K π = θ + π
3 k∈Z
When k = 1, θ is the minimum positive value
∴θ=2π−π
3=5π
Three
So the answer is: D
Shift the image of function y = sin2x to the right π 4 units and then 1 unit upward. The analytic expression of the image is () A. y=2cos2x B. y=2sin2x C. y=1+sin（2x+π 4） D. y=cos2x
Shift the image of function y = sin2x to the right π
Four units, the function y = sin2 (x - π) can be obtained
4) The image of = - cos2x,
Then the image is translated up by one unit, and the analytic expression of the image is y = - cos2x + 1 = 2sin2x,
Therefore, B
The value range of function y = sin2x + cos2x?
y=sin2x+cos2x
=Root 2 (Radix 2 * sin2x / 2 + Radix 2 * cos2x / 2)
=Root 2 (sin2x * cos45 ° + cos2x * cos45 °)
=Radical 2 * sin (2x + 45 °) ∈ [- radical 2, Radix 2]
I don't know now you understand (⊙_ ⊙)?
The value range of function y = sin2x-cos2x ,
The value range of function y = sin2x-cos2x
sin2x-cos2x=√2sin（2x+a）
The range is [- √ 2, √ 2]
What is the value range of the function y = sin2x + cos2x?
Sin2x + cos2x = √ 2 (√ 2 / 2sin2x + √ 2 / 2 * cos2x) = √ 2 (cospai / 4sin2x + sinpai / 4cos2x) = √ 2Sin (2x + Pai / 4), so... Formula: cosasinb + sinacosb = sin (a + b) y = sin2x + cos2x = √ 2 (sin45sin2x + cos45cos2x) = √ 2cos (45-2x) ∈ [- √ 2, √ 2]
Find the value range of the function y = sin2x-cos2x (x ∈ R),
y=sin2x-cos2x
=√2(√2/2sin2x-√2/2cos2x)
=√2(cosπ/4sin2x-sinπ/4cos2x)
=√2sin(2x-π/4)
So - √ 2
In order to get the image of function y = sin2x, the image of function y = cos2x can be shifted to the right by M unit length, then the maximum value of M is obtained
Pie / 4
2X right shift / 2 distance
Shift the image of the function y = sin2x + cos2x to the left π / 4
y=sin2x+cos2x=√2sin[2x+(π/4)]
Then, if we shift π / 4 to the left, we get: y = √ 2Sin [2 (x + π / 4) + (π / 4)] = √ 2Sin [2x + (3 π / 4)]
To get the image of the function y = sin2x, simply shift the image a of cos2x, shift π / 2 units to the left and π / 2 units to the right In order to get the image of the function y = sin2x, we only need to change the image of function cos2x A. Shift π / 2 units to the left B. Shift π / 2 units to the right C. Shift π / 4 units to the left D. Shift π / 4 units to the right Which one? Why
Cos2x = sin (2x + π / 2) = sin (2 * (x + π / 4)), select C
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