若0<α<π,则(1+sinα+cosα)(sinα2−cosα2)(2+2cosα)=_.

问题描述:

若0<α<π,则

(1+sinα+cosα)(sin
α
2
−cos
α
2
)
(2+2cosα)
=______.

∵0<α<π,
∴0<

α
2
π
2

∴cos
α
2
>0,
∴原式=
(2sin
α
2
cos
α
2
+2cos
2
α
2
)(sin
α
2
−cos
α
2
)
2cos2
α
2

=
2cos
α
2
(sin
α
2
+cos
α
2
)(sin
α
2
−cos
α
2
)
2cos
α
2

=sin2
α
2
-cos2
α
2

=-cosα.
故答案为:-cosα.