2sinx * cosx = sin2x, why?

2sinx * cosx = sin2x, why?

sin(x+x)=sinx*cosx+cosx*sinx=2sinx*cosx
Positive metaphysical formula of sum of two angles

Given (1 + TaNx) / (1-tanx) = 3 + 2 √ 2, find the value of (sin ^ 2x + √ 2sinx * cosx cos ^ 2x) / (sin 2x + cos ^ x) Given (1 + TaNx) / (1-tanx) = 3 + 2 √ 2, find the value of (sin ^ 2x + √ 2sinx * cosx cos ^ 2x) / (sin 2x + cos ^ x)

（1+tanx）/(1-tanx) =3+2√2 ,
The solution is TaNx = √ 2 / 2
(sin^2 x+√2sinx*cosx-cos^2 x)/(sin^2 x+2cos^2 x)
=(tan^2x+√2tanx-1)/(tan^2x+2)
=(1/2+1-1)/(1/2+2)
=1/5.

Solve the following trigonometric equation: (1) cosx-2sinx = 0 (2) SiNx = cos3x (3) sin ＾ 2x-3cos ＾ 2x = sin2x (4)3sin＾2x＋2sinx－1＝0(5)2sinxcosx＋sinx－cosx＝1

(1)cosx－2sinx＝0
tanx=1/2
x=kπ +arctan(1/2),k∈Z
(2)sinx＝cos3x
sin(2x-x)=cos(2x+x)
sin2xcosx-cos2xsinx=cos2xcosx-sin2xsinx
(sin2x-cos2x)(cosx+sinx)=0
tan2x=1 or tanx=-1
x=k(π/2)+π/8,or x=kπ+π/4,k∈Z
(3)sin＾2x－3cos＾2x＝sin2x
-1-2cos2x=sin2x
sin2x+2cos2x=-1
sin2x=-1/√5,cos2x=-2/V5
2x=(2k+1)π+arcsin(1/√5),k∈Z
(4)3sin＾2x＋2sinx－1＝0
sinx=1/3 or sinx=-1
x=kπ+(-1)^karcsin(1/3),or x=2kπ-π/2,k∈Z
(5)2sinxcosx＋sinx－cosx＝1
sinx-cosx=(sinx-cosx) ²
sinx-cosx=0 or sinx-cosx=1
x=kπ+π/4,or x=kπ+π/2,or x=kπ+π,k∈Z

Given TaNx = 2, find sin2x.cos2x.tan2x by using the triangular universal replacement formula

By the universal formula,
sin2x=（2tanx)/[1+(tanx)^2]
=2*2 / 12*2
=4/5
cos2x=[1-(tanx)^2]/[1+(tanx)^2]
=(1-2*2)/(1+2*2)
=-3/5
tan2x=（2tanx)/[1-(tanx)^2]
=(2*2)/(1-2*2)
=-3/4

If TaNx = 3, find the value of sin2x cos2x Please explain it in detail

sin2x-cos2x=2sinxcosx-(cos^2x-sin^2x)=2cos^2xtanx-(2cos^2x-1)
=6/(3^2+1)-2/(3^2+1)+1
=6/10-2/10+1
=7/5

Given TaNx = 2, find (sin2x + cos2x) / (cos2x sin2x)

Because TaNx = 2
So tan2x = 2tanx / [1 - (TaNx) ^ 2] = 2 * 2 / (1-2 ^ 2) = - 4 / 3
So (sin2x + cos2x) / (cos2x sin2x)
=[(sin2x/cos2x)+(cos2x)/(cos2x)]/[(cos2x)/(cos2x)-(sin2x)/(cos2x)]
=(tan2x+1)/(1-tan2x)
=[-4/3+1]/[1-(-4/3)]
=-1/7