In order to get the function y = cos (x + π / 3), we only need to shift the image of function y = SiNx (whether to shift 5 π / 6 unit length to the right) In order to get the function y = cos (x + π / 3), we only need to transform the image of function y = SiNx (is it 5 π / 6 unit length to the right)

In order to get the function y = cos (x + π / 3), we only need to shift the image of function y = SiNx (whether to shift 5 π / 6 unit length to the right) In order to get the function y = cos (x + π / 3), we only need to transform the image of function y = SiNx (is it 5 π / 6 unit length to the right)

Shift left 5 π / 6

Let f (x) = 2cosx (SiNx + cosx) - 1, shift the image of function f (x) to the left by a unit, and obtain the image of function y = g (x) Math homework help users 2017-10-06 report Use this app to check the operation efficiently and accurately!

（1） Simplification: F (x) = 2cosx (SiNx + cosx) - 1 = 2cosx * SiNx + 2cosx ^ 2-1 = sin2x + cos2x = sin (2x + pi / 4) (2) translation: F (x) - G (x) = sin [2 (x + a) + pi / 4] = sin (2x +...)

The image with function y = sin2x is moved to the left Pai / 6 units, and the function analytic formula of the translated image is

f(x)→f(x+π/6)
So y = sin2x → y = sin [2 (x + π / 6)] that is, y = sin [2x + π / 3]

All the points in the image of the function y = SiNx are shifted to the right π The abscissa of all points in the image is reduced to 1 The analytic formula is y = sin (ω x + φ), then () A. ω＝2，φ＝π Six B. ω＝2，φ＝−π Three C. ω＝1 2，φ＝π Six D. ω＝1 2，φ＝−π Twelve

All the points in the image of the function y = SiNx are shifted to the right π
Three units, y = sin (x - π)
3）
Then the abscissa of all points in the image is reduced to the original 1
2, y = sin (2x - π) is obtained
3）
∵ the analytic formula is y = sin (ω x + φ),
∴ω=2，φ=-π
3，
Therefore, B

Given the function y = SiNx, the abscissa of each point in the image is reduced to half of the original, and the ordinate remains unchanged, and then the image is shifted to the left Pai / 4 to obtain the analytic formula of the function

If the abscissa of each point in the image of the function y = SiNx is reduced to half of the original, and the ordinate remains unchanged, then y = sin2 (x + π / 4) = sin (2x + π / 2) = cos (2x) can be obtained by shifting the image to the left π / 4 to obtain y = sin2 (x + π / 4) = sin (2x + π / 2) = cos (2x)

Find the maximum and minimum value of the function y = SiNx cosx + sinxcos, X ∈ [0, Pai]

By (sinx-cosx) ^ 2 = 1-2 (sinx-cosx) ^ 2 = 1-2 (sinx-cosx) = [1 - (sinx-cosx) ^ 2] / 2, make sinx-cosx = t t = t = √ 2Sin (x-π / 4) - 1 ≤ t ≤ t ≤ √ 2Y = sinxcosx + sinx-cosx = (1-T ^ 2) / 2 + T = - T ^ 2 / 2 + T + 1 / 2 = - 1 / 2 (t-1) ^ 2 + 1 symmetric axis t = 1y on [- 1,1] on [- 1,1], monotonically increases monotonically on [1,1], and [1, √ 2] 2] 2] 2] 2 [1, √ 2] 2] 2 [1, √ 2] 2] 2 on monotone delivery

The minimum value of the function y = (SiNx / 2) + (2 / SiNx) in the interval (0, PAI) is

00,(2/sinx)>0
Y=(sinx/2)+(2/sinx)
=(sin^2x+4)/(2sinx)
2sinx≤2
SiNx = 1,2sinx max = 2
Y = (SiNx / 2) + (2 / SiNx) min = 2.5
Pay attention not to use the basic inequality, (SiNx / 2) = (2 / SiNx) does not hold!

Let x belong to (0, π), then the minimum value of the function y = SiNx / 2 + 2 / SiNx is

X belongs to (0, π)
Zero

Function y = SiNx + 1 The minimum value of SiNx (0 < x < π) is______ .

Let SiNx = t, ∵ 0 < x < π, ᙽ SiNx ∈ (0, 1], that is, t ∈ (0, 1]
The function y = SiNx + 1
sinx=t+1
t=1+1
T decreases monotonically at t ∈ (0, 1]
When t = 1, the function gets the minimum value of 2
∴y=sinx+1
The minimum value of SiNx (0 < x < π) is 2
So the answer is: 2

Given 0 < x < π, find the function y = SiNx + 2 The minimum value of SiNx is______ .

∵0＜x＜π，
∴0＜sinx≤1，
∴y=sinx+2
SiNx is at 0 < SiNx < 0
It decreases at 2:00,
When SiNx = 1,
∴y=sinx+2
SiNx gets the minimum value, Ymin = 1 + 2
1=3，
So the answer is: 3