(1)化简f(x),并求f(
);
(2)若0<α<π,f(α)+f(
)=0,求α.
(1)化简f(x),并求f(
);
(2)若0<α<π,f(α)+f(
)=0,求α.
解:
(1)f(x)==(2-sin2
)-(2-cos2
)=cos2
-sin2
=cosx,
∴f(
)=cos
π=cos
=
.
(2)由f(α)+f(
)=cosα+cos
=0,即2cos2
+cos
-1=0,
∴cos
=-1或cos
=
,∵0<α<π,
∴
=
,α=
π.