Y = SiNx + 1 divided by sinx-sinx + 2

Y = SiNx + 1 divided by sinx-sinx + 2

y= (sinx +1)/(sinx+1)(sinx-2)=1/(sinx-2) sinx∈[-1,1] y∈[-1,-1/3]

Find the range y = (1 + SiNx) / (2 + SiNx)

Y = (1 + SiNx) / (2 + SiNx) = (2 + SiNx - 1) / (2 + SiNx) = 1 - 1 / (2 + SiNx) - 1 ≤ SiNx ≤ 11 ≤ 2 + SiNx ≤ 31 / 3 ≤ 1 / (2 + SiNx) ≤ 1-1 ≤ - 1 / (2 + SiNx) ≤ - 1 / 30 ≤ 1 - 1 / (2 + SiNx) ≤ 2 / 3, so the range is [0, 2 / 3

Finding the range of y = 1-sinx / 1 + SiNx

It should be (1-sinx) / (1 + SiNx)
y=[2-(1+sinx)]/(1+sinx)
y=2/(1+sinx)-1
The value range of SiNx is [- 1,1]
And 1 + SiNx ≠ 0, so SiNx ≠ - 1
1+sinx∈(0,2]
2/(1+sinx)∈(∞,1]
y=2/(1+sinx)-1∈(∞,0]

Find the range of the following functions: (1) y = 4-3sin (x - π / 3) (2) y = cos ^ 2x SiNx

(1) The range of y = 4-3sin (x - π / 3) SiNx is [- 1,1], so the range of 4-3sin (x - π / 3) is [1,7] (2) y = cos ^ 2x SiNx = 1-2sin ^ 2x SiNx = - 2 (sin ^ 2x + 1 / 2sinx + 1 / 16) + 9 / 8 = - 2 (sinx-1 / 4) ^ 2 + 9 / 8sinx is [- 1,1], (sinx-1 / 4) ^ 2 is [0,25 / 16] C

How can y = 2Sin (2x - π / 3) be transformed online by y = SiNx

First, the abscissa of y = SiNx is changed to 1 / 2 of the original, and the ordinate is unchanged, and the image of y = sin2x is obtained
Then, y = sin (2x - π / 3) is obtained by translating π / 6 of y = sin 2x to the right
By changing the ordinate of y = sin (2x - π / 3) to 2 times of the original and keeping the abscissa unchanged, we get y = 2Sin (2x - π / 3)

It shows how the image of y = 2Sin (2x + π) can be transformed from the image of y = SiNx

Step 1: y = SiNx shifts π units to the left to get y = sin (x + π) [translation transformation]
Step 2: y = sin (x + π) ordinate unchanged, abscissa changed to 1 / 2 times of the original, y = sin (2x + π) [lateral expansion transformation]
The third step: y = sin (2x + π) abscissa unchanged, ordinate changed to twice the original, y = 2Sin (2x + π) [longitudinal expansion transformation]

Explain how the image of y = - 2Sin (2x - π / 6) + 1 is transformed from the image of y = SiNx?
The process should be complete

In general, there are two ways to get: Method 1: first, y = SiN x moves π / 6 units to the right to get y = sin (x - π / 6) [translation transformation] second, y = sin (x - π / 6) ordinate remains unchanged, abscissa becomes 1 / 2 times of the original to get y = sin (2x - π / 6) [lateral expansion transformation] third, y = sin (2x -...)

Transform y = 2Sin (2x + Π / 3) + 2, X ∈ r to y = SiNx (two methods)

Method (1): translate the image of y = 2Sin (2x + Π / 3) + 2 down two units to get the image of y = 2Sin (2x + Π / 3), then keep the abscissa unchanged, reduce the ordinate to the original 1 / 2 to get the image of y = sin (2X + Π / 3), then move the image of y = sin (2x + Π / 3) right to Π / 3 single bits to get the image of y = sin2x, and keep the ordinate unchanged

sin(x+△x)-sinx=2sin△x/2cos(x+△x/2)
Why is the master
sin(x+△x)-sinx=2sin△x/2 cos(x+△x/2)
The condition is to prove that SiNx is a continuous function on R!

sin(x+△x)-sinβ=2sin(△x/2) * cos(x+△x/2)
The formula of sum difference product
sinα-sinβ
=2cos[(α+β)/2]sin[(α-β)/2]
In addition, α = x + △ x, β = X
The same result: sin (x + △ x) - sin β = 2Sin (△ X / 2) * cos (x + △ X / 2)

High school mathematics SiNx + root 3cosx = 1, written as 2Sin (x + π / 3) = 1, how to calculate π / 3?

Original formula: SiNx + √ 3cosx = 1
2((1/2)*sinx+(√3/2)cosx)=1
2(sinx*cos(π/3)+cosx*sin(π/3))=1
2sin(x+π/3)=1
(because cos (π / 3) = 1 / 2, sin (π / 3) = 3 / 2)