已知 cos(x-
(I)求sinx的值; (Ⅱ)求sin(2x+
|
(I)∵cos(x-
)=π 4
,x∈(2 10
,π).π 2
∴
(sinx+cosx)=2 2
;2 10
⇒sinx+conx=
⇒cosx=1 5
-sinx;1 5
代入sin2x+cos2x=1解得sinx=
,cosx=-4 5
.3 5
(Ⅱ)∵sinx=
,cosx=-4 5
.3 5
∴sin2x=2sinxcosx=-
;12 25
cos2x=2cos2x-1=-
.7 25
∴sin(2x+
)=sin2xcosπ 3
+cos2xsinπ 3 π 3
=-
×12 25
+(-1 2
)×7 25 3 2 ( ) ( )
=-
.12+7 3 50