Let the function f (x) = (SIN) ω x+cos ω x) The minimum positive period of ^ + 2cos ^ Wx (W > 0) is 2 π / 3 ω If the image of function y = g (x) is composed of y = f (x)

fx=sin^wx+cos^wx+2sinwxcoswx+2*(1+cos2wx)/2
=1+sin2wx+cos2wx+1
=sqrt(2)Sin(2wx+π/2)+2
Because t = 2 π / W, 2 π / 2W = 2 π / 3, so π = 3 / 2

Find the derivative of the following function (1) y = x ^ 2 + X + 1 (2) y = cos (x + 3)

y '=(x ²+ x+1) '
=2x+1
y '=-sin(x+3)

In order to get the image of the function y = cos (x + π / 3), just put the image of the function y = SiNx () A shifts π / 6 length units to the left B shift π / 6 length units to the right C shifts 5 π / 6 length units to the left D shifts 5 π / 6 length units to the right [the image of trigonometric function is always incomprehensible. Please explain it in detail] Know left plus right minus, up plus down minus But I don't know how cos becomes sin

It is treated with the constant equation sin (π / 2-x) = cos x = cos (- x)
cos(x+π/3)=cos(-x-π/3)=sin[π/2-(-x-π/3)]=sin(x+5π/6)
Shift 5 π / 6 length units to the left, select C

In order to get the image of function y = cos (x - π / 3), how can the image of function y = SiNx be moved?

y=sinx=cos(π/2- x)= cos( x-π/2)
Move it to the left π / 6 to get
y= cos( x-π/2+π/6)= cos(x-π/3)
Therefore, the image of function y = cos (x - π / 3) can be obtained by moving the image of function y = SiNx to the left π / 6

Minimum positive period of function y = cos (SiNx)

Minimum positive period T = π
y(x+π)=cos(sin(x+π))=cos(-sinx))=cos(sinx)=y(x)
There is no period smaller than π, because when x = 0, y (0) = cos0 = 1
When x is between (0, π), 0

Let y = sin2x / (1 + x ^ 2), find dy / DX,

dy/dx=[(sin2x) '(1+x ²)- sin2x·(1+x ²)']/ (1+x ²)²
=[2cos2x·(1+x ²)- 2x·sin2x]/(1+x ²)²

Let the function f (x) = (SIN) ω x+cos ω x) The minimum positive period of ^ + 2cos ^ Wx (W > 0) is 2 π / 3 ω If the image of function y = g (x) is composed of y = f (x)