化简:(1)sin[α+(2n+1)π]+sin[α-(2n+1)π]sin(α+2nπ)•cos(α-2nπ)(2)1-cos4α-sin4α1-cos6α-sin6α.-数学

题目简介

化简:(1)sin[α+(2n+1)π]+sin[α-(2n+1)π]sin(α+2nπ)•cos(α-2nπ)(2)1-cos4α-sin4α1-cos6α-sin6α.-数学

题目详情

化简:
(1)
sin[α+(2n+1)π]+sin[α-(2n+1)π]
sin(α+2nπ)•cos(α-2nπ)

(2)
1-cos4α-sin4α
1-cos6α-sin6α
题型:解答题难度:中档来源:不详

答案

原式=
sin(2nπ+π+α)+sin(-2nπ-π+α)
sin(2nπ+α)•cos(-2nπ+α)
=
sin(π+α)+sin(-π+α)
sinα•cosα

=
-sinα-sin(π-α)
sinα•cosα
=class="stub"-2sinα
sinα•cosα
=-class="stub"2
cosα

2)原式=
(cos2α+sin2α)2-cos4α-sin4α
(cos2α+sin2α)3-cos6α-sin6α
=
2cos2α•sin2α
3cos2αsin2α(cos2α+sin2α)
=class="stub"2
3

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