已知
(1)求
(2)求
|
∵
<α<π,sin(π-α)=sinα=π 2
,4 5
∴cosα=-
=-1-sin2α
,故tanα=3 5
=-sinα cosα
,4 3
(Ⅰ)由诱导公式可得sin(2π+α)tan(π-α)cos(-π-α) sin(
-α)cos(3π 2
+α)π 2
=
=-tanα=sinα•(-tanα)•(-cosα) -cosα•(-sinα)
;4 3
(Ⅱ)由诱导公式可得sin(π-α)+cos(-α) tan(π+α)
=
=sinα+cosα tanα
=-
-4 5 3 5 - 4 3 3 20