问题 解答题
化简求值:
(1)
sin(2π-α)sin(π+α)cos(-π-α)
cos(
π
2
+α)sin(3π-α)cos(π-α)

(2)log2.56.25+lg
1
100
+ln
e
+21+log23
答案

(1)∵sin(2π-α)=-sinα,sin(π+α)=-sinα,cos(-π-α)=-cosα

cos(

π
2
+α)=-sinα,sin(3π-α)=sinα,cos(π-α)=-cosα

sin(2π-α)sin(π+α)cos(-π-α)
cos(
π
2
+α)sin(3π-α)cos(π-α)

=

-sinα•(-sinα)•(-cosα)
-sinα•sinα•(-cosα)
=-1

(2)∵6.25=2.52

1
100
=10-2
e
=e
1
2
,1+log23=log26

log2.56.25+lg

1
100
+ln
e
+21+log23

=log2.52.52+lg10-2+lne

1
2
+2log26=2+(-2)+
1
2
+6=
13
2

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