试题:
化简
sin(2π-α)cos(π+α)cos(
π
2
+α)cos(
11
2
π-α)
cos(π-α)sin(3π-α)sin(-π-α)sin(
9
2
π+α)

答案:

我来补答
sin(2π-α)cos(π+α)cos(
π
2
+α)cos(
11
2
π-α)
cos(π-α)sin(3π-α)sin(-π-α)sin(
9
2
π+α)
=
-sinα•(-cosα)(-sinα)(-sinα)
-cosα•sinα•sinα•cosα
=-tanα.
 
 
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