Sin2x-cos2x-1 = √ 2 sin (2x - π / 4) how does this work?

Sin2x-cos2x-1 = √ 2 sin (2x - π / 4) how does this work?

sin2x-cos2x-1
=√2(sin2x×√2/2-cos2x×√2/2)-1
=√2sin(2x-π/4)-1
If there is anything you don't understand, you can ask,

Given that 2Sin? X-cos? X + sinxcosx-4sinx + 2cosx = 0, find (sin2x + 1) / (1 + cos2x + sin2x)

（2sinx-cosx）（sinx+cosx）-2（2sinx-cosx）=0
（2sinx-cosx）（sinx+cosx-2）=0
Because (SiNx + cosx-2) < 0
So (2sinx cosx) = 0
2sinx=cosx
tanx=1/2
(sin2x+1)/(1+cos2x+sin2x)
=The denominator of the denominator is 2xs + s22xs + s22xs
=(2tanx+tan^2x+1)/(2+2tanx)
=(1+1/4+1)/(2+1)
=3/4

Confirmation: cos2x = cos? X-sin? X

According to the formula of the sum of two angles of trigonometric function:
cos(α+β)=cosα·cosβ-sinα·sinβ
cos2x=cos(x+x)
=cosxcosx-sinxsinx
=cos²x-sin²x

Given the function f (x) = sin (2x + π / 6) + sin (2x - π / 6) + cos2x + A, (1) find the minimum positive period and monotonicity of the function

The first problem is: F (x) = sin2xcos (π / 6) + cos2xsin (π / 6) + sin2xcos (π / 6) - cos2xsin (π / 6) - cos2x + a = 2sin2xcos (π / 6) - cos2x + a = 2 [sin2xcos (π / 6) - cos2xsin (π / 6)] + a = 2Sin (2x - π / 6) + a

The minimum positive period of the function y = sin (π / 6-2x) + cos2x is

y=sin(π/6-2x)+cos2x
=1 / 2 * cos2x - (radical 3) / 2 * sin2x + cos2x
=3 / 2 * cos2x - (radical 3) / 2 * sin2x
=(radical 3) * [(radical 3) / 2 * cos2x-1 / 2 * sin2x]
=(radical 3) * cos (2x + π / 6)
So w = 2, the minimum positive period is t = 2 π / w = π

Given the function f (x) = sin (2x + Pai / 6) + cos2x + 1, find the minimum positive period

F (x) = sin2x * (root 3 / 2) + cos2x * (1 / 2) + cos2x + 1
=Root 3 / 2 * sin2x + 3 / 2cos2x + 1
=Root 3 (1 / 2 * sin2x + root 3 / 2 * cos2x) + 1
=Root 3 * sin (2x + pi / 3) + 1
Therefore, the minimum period is 2pi / 2 = Pi
Because we can't transfer the pictures, the writing is not standard!

The function y = sin (π The minimum positive period of 3 − 2x) + cos2x is______ .

∵f（x）=sin（ π
3−2x）+cos2x=
Three
2cos2x-1
2sin2x+cos2x=（
Three
2+1）cos2x-1
2sin2x
=
2+
3sin（2x+θ）
∴T=2π
2=π
So the answer is: π

The known function f (x) = sin (2x + π) 3)+sin(2x−π 3)+cos2x (1) Find the minimum positive period of function f (x); (2) The image of the function f (x) along the vector m＝(−3π 8,2) the image of function g (x) is obtained by translation, and the monotone decreasing interval of function g (x) on X ∈ [0, π] is obtained

f(x)＝2sin2xcosπ
3+cos2x＝sin2x+cos2x=
2sin(2x+π
4)… (4 points)
(1) The minimum positive period of function f (x) is 2 π
2＝π… (6 points)
(2) G (x) = f (x + 3 π)
8）+2=
2sin（2x+3π
4+π
4）+2=−
2sin2x+2… (8 points)
∵0≤x≤π∴0≤2x≤2π
Monotonically decreasing by G (x) on [0, π]
∴0≤2x≤π
2, or 3 π
2≤2x≤2π
∵0≤x≤π
4, or 3 π
4≤2x≤π… (11 points)
So the monotone decreasing interval of function f (x) is [0, π] and [3 π]
4，π]… (12 points)

The function f (x) = sin (2x + π / 6) + sin (2x - π / 6) - cos2x + a (a is a real number and belongs to R) (1) Find the minimum positive period (2) Finding monotone increasing interval of function (3) If x belongs to [0, Pai / 2], the minimum value of F (x) is - 2

First question:
f（x）＝sin2xcos（π/6）＋cos2xsin（π/6）＋sin2xcos（π/6）－cos2xsin（π/6）－cos2x＋a
＝2sin2xcos（π/6）－cos2x＋a＝2［sin2xcos（π/6）－cos2xsin（π/6）］＋a
＝2sin（2x－π/6）＋a.
The minimum periodic function of π (2) is π
Second question:
When 2K π - π / 2 ≤ 2 π - π / 6 ≤ 2K π + π / 2, f (x) increases monotonically
From 2K π - π / 2 ≤ 2x - π / 6 ≤ 2K π + π / 2, 2K π - 3 π / 6 + π / 6 ≤ 2x ≤ 2K π + 3 π / 6 + π / 6 is obtained,
∴2kπ－2π/6≦2x≦2kπ＋4π/6,∴kπ－π/6≦x≦kπ＋π/3.
In other words, the monotone increasing interval of function f (x) is [K π - π / 6, K π + π / 3], where k is an integer
Third question:
∵0≦x≦π/2 ,∴0≦2x≦π,∴－π/6≦2x－π/6≦π－π/6,
The minimum value of F (x) is 2Sin (- π / 6) + a = - 1 + a = - 2, a = - 1

The known function f (x) = sin (2x + π / 6) - cos2x Find the minimum positive period and monotone increasing interval of F (x) 2. Find the minimum and maximum values of F (x) on [0, π / 2] and the corresponding x values 3 if the function f (x) satisfies the equation f (x) = a (0 < a < 1), find the sum of all real roots in [0,2 π]

f(x)=sin(2x+π/6)-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-π/3)
1. The minimum positive period = 2 π / 2 = π
Monotonically increasing interval 2K π - π / 2 ≤ 2x - π / 3 ≤ 2K π + π / 2
X ∈ [K π - π / 12, K π + 5 π / 12] is obtained
When 2.2x - π / 3 = π / 2 x = 5 π / 12, the maximum value of F (x) is 1
When 2x - π / 3 = 3 π / 2 x = 11 π / 12, the minimum of F (x) = - 1
3.a>0 sin(2x-π/3)>0
In [0,2 π]
sin(2x-π/3)=a
2x-π/3=arcsina x=π/6+(1/2)arcsina
Or 2x - π / 3 = π - arcsina x = 2 π / 3 - (1 / 2) arcsina
So the sum of all real roots is π / 6 + (1 / 2) arcsina + 2 π / 3 - (1 / 2) arcsina
=5π/6