sin^2x+cos2x=1 Process of solving equation

sin^2x+cos2x=1 Process of solving equation

cos2x=cos^2x-sin^2x
The original formula = sin ^ 2x + cos ^ 2x - Sin ^ 2x = cos ^ 2x = 1
Cosx = 1 or - 1
X = k * π K is an integer

=How does 1 / 2 * (sin2x cos2x) - 1 / 2 become = √ 2 / 2 * sin (2x + π / 4) - 1 / 2

1 / 2 * (sin2x cos2x) - 1 / 2 extracted in brackets √ 2
=√2/2*(√2/2xin2x-√2/2cos2x)-1/2
=√2/2*(sin2xcosπ/4-cos2xsinπ/4)-1/2
=√2/2*sin(2x-π/4)-1/2
Your question is = √ 2 / 2 * sin (2x + π / 4) - 1 / 2 is wrong

How does f (x) =sin (2x+ π /3) become g (x) =cos2x Find specific steps, what formula to use, etc

Can't f (x) = sin (2x + π / 3) be cos2x
It can be changed into cos (2x + φ) form
f(x)=sin(2x+π/3)
=sin[2x+(π/2-π/6)]
=Sin [π / 2 + (2x - π / 6)] [induction formula: sin (π / 2 + α) = cos α]
=cos(2x-π/6)

How does f (x) = sin (2x + 3) become f (x) = - cos2x

f(x)=sin(2x+3π/2)=sin(2x+3π/2-2π)
=sin(2x-π/2)=-sin(π/2-2x)=-cos2x

How much is root 3 * 1 + cos2x / 2, and why is root 3 / 2cos2x + Radix 3 / 2,

√3*(1+cos2x/2)
=√3+√3/2cos2x
=√3/2cos2x +√3
Not root 3 / 2cos2x + radical 3 / 2

Given the function f (x) = 1 / 2sin2x - (radical 3 / 2) cos2x + 1, if f (x) > = log2t holds, find the range of T

f(x)=1/2sin2x-（√3/2）cos2x+1
=sin(2x-π/3)+1
Minimum = - 1 + 1 = 0
F (x) > = log (2) t always holds
∴0>=log(2)t
∴0

Find the period y = root (1-cos2x) + radical (1 + cos2x)

According to the formula of double angle angle, there are cos2x = (cosx) ^ 2 - (SiNx) ^ 2 = 1-2 (SiNx) ^ 2 = 2 (cosx) ^ 2 = 2 (cosx) ^ 2-1 √ (1-cos2x) = √ 2 * | SiNx |,, (1 + cos2x) = √ 2 * | cosx | y = √ 2 * || SiNx | (2 | cosx | | the periods of both SiNx and cosx are 2 π, and after adding absolute value number, it is equivalent to the corresponding to the absolute value number after adding the absolute value number, it is equivalent to the corresponding to the period of the period is 2 π, and after adding the absolute value number, it is equivalent put the image at the bottom of the x-axis up

The result of root 2 + 2cos2x simplification is

2+2cos2x
=2+2(2cos²x-1)
=4cos²x
=(2cosx)²
So the original formula = 2|cosx|

Simplify √ 2 + 2sin2x ("√" stands for radical)

First of all, look inside the radical
2+2sin2x=2+4sinxcosx=2(sinx^2+cosx^2+2sinxcosx)=2(sinx+cosx)^2
So root 2 + 2sin2x = Radix 2 * | SiNx + cosx | = Radix 2 * Radix 2 * | sin (x + 45 degrees)|
=2 * | sin (x + 45 degrees)|
be
When - 45 + 360K

2sin2x-3/2cos2x+3/2 simplification

The extraction coefficient of the first two terms of the auxiliary angle formula is √ [2] + (- 3 / 2) 2] = 5 / 2  2sin2x-3 / 2cos2x + 3 / 2 = 5 / 2 * (4 / 5 * sin2x-3 / 5 * cos2x) + 3 / 2 let sin α = 3 / 5, cos α = 4 / 5 (α is a constant value, is an acute angle)  the original formula = (5 / 2sin2x - α) + 3 / 2