It is known that 0, , and 2sin2 sin 1 cos is verified

So. The original inequality is.

4sin cos sin (1-cos) because (0, ], so sin is greater than or equal to 0

So the inequality is reduced to.

4cosα≤1/(1-cosα)

Only proof is required.

cosα(1-cosα)≤1/4

If cos 0

cosα(1-cosα)≤0

Established cos (1-cos) 1 4

If cos 0

then cos (1-cos) 2 = 1 4

Also established. Certification.

Certificate: If the left and right are equal.

If (0, ).

sinα>0,1-cosα>0;

2sin2α-sinα/(1-cosα)=sinα/(1-cosα)*4(cosα)^2+4cosα-1]

2cos -1) 2*sin (1-cos) 0 In summary. 2sin2 sin (1-cos).

4sinxcosx≤sinα/(1-cosα)4sinx(cosx-1/(1-cosx))≤0sinx>=0

Then cosx-1 (1-cosx) 0

cosx-cos^2x-1)(1-cosx)≤01-cosx>0

cosx-cos^2x-1≤0

cosx is between (1,-1).

The parabola opens downwards and has no intersection with the x-axis.

Then it holds. Hence 2sin2 sin (1-cos).

The original formula becomes 2sin2 -sin (1-cos) 02sin cos -sin (1-cos ) 0sin >=0 divided by sin 2 on both sides

2cosα-1/(1-cosα)≤0

1-cos >=0 multiply 1-cos on both sides at the same time

2cosα(1-cosα)-1≤0

Left = 2cos -2cos *2cos -1-[2(cos -1 2) 2+1 2].

This equation sub 0 original formula is proved.

1 All sins 1-cos = 2sin( 2)cos( 2) 2sin( 2) 2=cos( 2) sin( 2) = cot( 2), let k = cot( 2), have:

tan( 2)=1 k,tan =2tan( 2) [1-tan( 2) 2]=2k (k 2-1),sin 2=tan 2 (tan 2+1),cos 2=1 (tan 2+1),sin2 2=(2sin cos) 2=4*sin 2cos 2=4*tan 2 (tan 2+1) 2, (0, 2), 2 (0, )sin2 >0,tan >0,k=1 tan >0,2sin2 =2 [4*tan 2 (tan 2+1) 2]=4*tan (tan 2+1)=8*(k 3-k) (k 4+2k 2+1),2sin2 -sin (1-cos )=8*(k 3-k) (k 4+2k 2+1)-k=(-k 5+6k 3-9k) (k 4+2k 2+1)=-k*(k 2-3) 2 (k 2+1) 2, Since k>0,-k*(k2-3)2(k2+1) 2<0, 2sin2 -sin (1-cos) < 0,2sin2 0, so 2sin2 -sin (1-cos) >0, in summary, 0, 2], 2sin2 sin (1-cos).

Proof: If left and right are equal.

If (0, ).

sinα>0,1-cosα>0;

2sin2α-sinα/(1-cosα)=sinα/(1-cosα)*4(cosα)^2+4cosα-1]

2cos -1) 2*sin (1-cos) 0 In summary.

2sin2α-sinα/(1-cosα)=4sinαcosα-sinα/(1-cosα)

(1-cos )4sin cos -sin ) (1-cos )=sin (4cos -4cos 2-1) (1-cos )=-sin(2cos -1) 2 (1-cos) because belongs to (0, ).

So sin >0, cos belongs to (-1,1), 1-cos >0 and because (2cos -1) 2>=0

So-sin(2cos -1) 2 (1-cos)<0 so 2sin2

（1）α=x

0,0(sinx) 2,cosx>(cosx) 2sinx+cosx>(sinx) 2+(cosx) 2=1(2)

2sin2 hail sin (1-cos) proves: if the left and right are equal to the state He Fan.

If pat (0, ).

sinα>0,1-cosα>0;

2sin2α-sinα/(1-cosα)=sinα/(1-cosα)*4(cosα)^2+4cosα-1]

2cos -1) 2*sin (1-cos) 0 In summary. 2sin2α≤sinα/(1-cosα)

(0, read hail early).

sinα>0,1-cosα>0;

2sin2 -sin (1-cos) = sin (1-cos) *4 (cos) 2+4cos -1].

2cos -1) 2*sin (1-cos) 0 In summary, 2sin2< sin (1-cos) < p>

Cartesian coordinate plane.

Let the terminal edge of the acute angle intersect the unit circle and the point p,a(1,0) sinusoidal mp sin

Tangent at= tan

Apparently mp1 2 sin

sin to sum up. sinα<αtanα