Because 0

So,

2 sin2a-[ Sina /( 1-cosa)]

=[4 Sina cosa( 1-cosa)-Sina]/( 1-cosa)

=-[sina/( 1-cosa)](2cosa- 1)^2

= & lt0

Therefore,

2 sin2a = < Sina /( 1-cosa).

2.( 1) Derive f ′ (x) = ax ㏑ a+[x+1-(x-2)]/(x+1)? = a xlna+3/(x+ 1), because x >；; 1，a & gt 1,a^x>； 0，lna & gt0，3/(x+ 1)？ & gt0。 So f' (x) >: 0 means that f(x) is increasing function on (-1, +∞).

(2) when x <-1, ax >; 0，(x-2)/(x+ 1)>0

a^x+(x-2)/(x+ 1)>； 0, f(x)=0 has no root, so there is no negative real root.

When-1

Because of a> 1, an X.

(x-2)/(x+ 1)= 1-3/(x+ 1)& lt； -2

a^x+(x-2)/(x+ 1)<； - 1 & lt； 0

F(x)=0 has no root, that is, there is no negative root.

x≦- 1

So the equation f(x)=0 has no negative root.