化简:(Ⅰ)sin(α-2π)cos(α+π)tan(α-99π)cos(π-α)sin(3π-α)sin(-α-π);(Ⅱ)sin(nπ+α)cos(nπ-α)(n∈Z).-数学

题目简介

化简:(Ⅰ)sin(α-2π)cos(α+π)tan(α-99π)cos(π-α)sin(3π-α)sin(-α-π);(Ⅱ)sin(nπ+α)cos(nπ-α)(n∈Z).-数学

题目详情

化简:(Ⅰ)
sin(α-2π)cos(α+π)tan(α-99π)
cos(π-α)sin(3π-α)sin(-α-π)
;    (Ⅱ)
sin(nπ+α)
cos(nπ-α)
  (n∈Z)
题型:解答题难度:中档来源:不详

答案

(Ⅰ)原式=
sinα•(-cosα)•tanα
-cosα•sinα•sinα

=class="stub"tanα
sinα
=class="stub"1
cosα

(Ⅱ)当n=2k,k∈Z时原式=
sin(2kπ+α)
cos(2kπ-α)
=class="stub"sinα
cosα
=tanα

当n=2k+1,k∈Z时原式=
sin(2kπ+π+α)
cos(2kπ+π-α)
=class="stub"-sinα
-cosα
=tanα

∴当n∈Z时原式=tanα

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