(1)化简sin(2π-α)cos(π+α)cos(π-α)sin(3π-α)sin(-α-π);(2)求值:3tan12°-3sin12°(4cos212°-2).-数学

题目简介

(1)化简sin(2π-α)cos(π+α)cos(π-α)sin(3π-α)sin(-α-π);(2)求值:3tan12°-3sin12°(4cos212°-2).-数学

题目详情

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

(2)求值:
3
tan12°-3
sin12°(4cos212°-2)
题型:解答题难度:中档来源:不详

答案

(1)
sin(2π-α)cos(π+α)
cos(π-α)sin(3π-α)sin(-α-π)
=
-sinα•(-cosα)
-cosα•sinα•sinα
=-class="stub"1
sinα

(2)
3
tan12°-3
sin12°(4cos212°-2)

=
3
(class="stub"sin12°
cos12°
-
3
)
2sin12°cos24°

=
2
3
(class="stub"1
2
sin12°-
3
2
cos12°)
2sin12°•cos12°•cos24°

=
2
3
sin(12°-60°)
sin24°•cos24°

=-
2
3
sin48°
class="stub"1
2
sin48°

=-4
3

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