化简sin1度6次方 sin2度6次方 ····sin89度6次方=
答案:1 悬赏:30
解决时间 2021-02-27 03:36
- 提问者网友:末路
- 2021-02-26 08:02
化简sin1度6次方 sin2度6次方 ····sin89度6次方=
最佳答案
- 二级知识专家网友:春色三分
- 2021-02-26 09:38
先求(sinα)^6+(cosα)^6 的值
(sinα)^6+(cosα)^6
=[(sinα)^2+(cosα)^2][(sinα)^4-(sinα)^2(cosα)^2+(cosα)^4]
=1[(sin²α+cos²α)²-3sin²αcos²α]
=1-3/4(4sinαcosα)²
=1-3/4(sin²2α)
=1-3/8(1-cos4α)
=5/8+3/8*cos4α
∴(sin1º)^6+(sin89º)^6
=(sin1º)^6+(cos1º)^6
=5/8+3/8 cos4º
即(sin1º)^6+(sin89º)^6=5/8+3/8 cos4º
同理:(sin2º)^6+(sin88º)^6=5/8+3/8 cos8º
(sin3º)^6+(sin87º)^6=5/8+3/8 cos12º
.........................................................
(sin44º)^6+(sin46º)^6=5/8+3/8 cos176º
(sin45º)^6=(√2/2)^6=1/8
将上面45个等式相加:
即
sin1度6次方+ sin2度6次方 ····sin89度6次方
=5/8*44+1/8+3/8( cos4º+cos8º+cos12º+....+cos168º+cos172º+cos176º)
=221/8
[注意:cos4º+cos176º=0
cos8º+cos172º=0
..........................
∴cos4º+cos8º+cos12º+....+cos168º+cos172º+cos176º=0]
(sinα)^6+(cosα)^6
=[(sinα)^2+(cosα)^2][(sinα)^4-(sinα)^2(cosα)^2+(cosα)^4]
=1[(sin²α+cos²α)²-3sin²αcos²α]
=1-3/4(4sinαcosα)²
=1-3/4(sin²2α)
=1-3/8(1-cos4α)
=5/8+3/8*cos4α
∴(sin1º)^6+(sin89º)^6
=(sin1º)^6+(cos1º)^6
=5/8+3/8 cos4º
即(sin1º)^6+(sin89º)^6=5/8+3/8 cos4º
同理:(sin2º)^6+(sin88º)^6=5/8+3/8 cos8º
(sin3º)^6+(sin87º)^6=5/8+3/8 cos12º
.........................................................
(sin44º)^6+(sin46º)^6=5/8+3/8 cos176º
(sin45º)^6=(√2/2)^6=1/8
将上面45个等式相加:
即
sin1度6次方+ sin2度6次方 ····sin89度6次方
=5/8*44+1/8+3/8( cos4º+cos8º+cos12º+....+cos168º+cos172º+cos176º)
=221/8
[注意:cos4º+cos176º=0
cos8º+cos172º=0
..........................
∴cos4º+cos8º+cos12º+....+cos168º+cos172º+cos176º=0]
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