Find the maximum, minimum and minimum period of y = SiNx sin (x + 45 degrees)

Find the maximum, minimum and minimum period of y = SiNx sin (x + 45 degrees)

The original function can be changed into:
y=sinX-(cos45sinX+sin45cosX)
The results show that y = (1-cos45) sinx-sin45cosx

Y = SiNx sin (x + π / 4) maximum value, minimum value period

y=sinx-sin（x+π/4)
=2cos[(2x+π/4)/2]sin[（x-x-π/4）/2]
=-2sinπ/8 cos(x+π/8)
therefore
Maximum = 2Sin π / 8
Minimum = - 2Sin π / 8
Period = 2 π

Y = the maximum of COS (x + 3) - sin (x + 3 / 3). Minimum, period?

y=cos(x＋π/3)－sin(x＋π/3)
=√2[(√2/2)cos(x＋π/3)－(√2/2)sin(x＋π/3)]
=√2cos[(x＋π/3)＋(π/4)]
=√2cos(x＋7π/12)
The maximum value of this function is √ 2, the minimum value is - 2, and the period is 2 π

Y = SiNx sin (x + π / 4) max min)

y=sinx-sin(x+π/4)
=sinx-[sinxcos(π/4)+cosxsin(π/4)]
=sinx-[(√2/2)sinx +(√2/2)cosx]
=(1- √2/2)sinx -(√2/2)cosx
=√ [(1 - √ 2 / 2) 2 + (- √ 2 / 2)] sin (x + a), where Tana = - (√ 2 / 2) / (1 - √ 2 / 2)
=√(2 -√2)sin(x+a)
When sin (x + a) = 1, y has a maximum value ymax = √ (2 - √ 2)
When sin (x + a) = - 1, y has a minimum value, Ymin = - √ (2 - √ 2)

The relationship between y = SiNx and y = sin ω x

The amplitude is the same, but the period is different

y＝sinx×sin(x＋π/6)

Y = SiNx × sin (x + π / 6) = SiNx (√ 3 / 2sinx + 1 / 2cosx) = √ 3 / 2Sin ^ 2x + 1 / 2sinxcosx = √ 3 / 4 (1-cos2x) + 1 / 4sin2x = 1 / 2Sin (2x - π / 3) + √ 3 / 4, what is the minimum positive period T = π, maximum y = (2 + √ 3) / 4, minimum y = (- 2 + √ 3) / 4K π - π / 12

Find the minimum positive period y = | SiNx | y = sin | X| Can the combination of nonnumeral forms be solved?

Of course, obviously
|sin（x+π）|=|-sinx|=|sinx|
So the minimum positive period is π
Sin | x | is the whole definition domain, which is aperiodic function
In (0, + ∞), is a periodic function
sin（x+2π）=sinx
So the minimum positive period is 2 π
Similarly, in (- ∞, 0), it is a periodic function
sin（-x+2π）=sin-x
So the minimum positive period is 2 π

The image of function y = SiNx is shifted to the left by π / 3 unit length to get image C1. Then, the abscissa of all points on C1 is changed to twice of the original, and the ordinate remains unchanged. The analytic formula of image C2, C2 is obtained?

The image of function y = SiNx is shifted to the left by π / 3 unit length to obtain image C1
The analytic formula of C1 is y = sin (x + π / 3)
The abscissa of all points on C1 is changed to twice of the original, and the ordinate remains unchanged, and the image C2 is obtained
Then the analytic formula of C2 is y = sin (x / 2 + π / 6)

The image of the function y = f (x) SiNx is shifted to the right by π / 4 unit length, and then the symmetric transformation about X axis is made to obtain the image of function y = - cos2x, and f (x) is obtained I know it turns out to be f (x) SiNx = - sin2x, and then what?

Y = - cos2x is symmetrically transformed with respect to X axis to obtain y = cos2x
Then shift π / 4 units to the left to obtain y = cos (2x + π / 2) = - sin2x
So f (x) SiNx = - sin2x = - 2sinxcosx
So f (x) = - cosx domain is SiNx is not 0
X is not k π (k is an integer)
So f (x) = - cosx x x is not k π (k is an integer)

The image of function y = cos2x is shifted to the right π / 4 unit length to get the image of function y = f (x) * SiNx, then f (x) can be

y=cos[2(x-π/4)]
=cos(2x-π/2)
=sin2x
=2sinxcosx
f(x)=2cosx