sin平方(π/6-x)怎么化简
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解决时间 2021-04-29 07:00
- 提问者网友：两耳就是菩提
- 2021-04-29 00:29
sin平方(π/6-x)怎么化简
最佳答案
- 五星知识达人网友：过活
- 2021-04-29 01:05
(1/2)(1 + cos2x) + (cosx)[cosxcos(π/3) – sinxsin(π/3)] + (1/2)[1– cos(π/3 – 2x)]
= 1/2 + (1/2)cos2x+ (1/2)cos 2 x– (√3/2)cosxsinx+ 1/2 – (1/2)cos(π/3 – 2x)
= 1 +(1/2)cos2x + (1/2)*(1/2)(1 + cos2x) – (√3/4)sin2x – (1/2)[cos(π/3)cos2x+ sin(π/3)sin2x]
= 5/4 + (3/4)cos2x– (√3/4)sin2x – (1/4)cos2x – (√3/4)sin2x
= 5/4 + (1/2)cos2x – (√3/2)sin2x
= 5/4 + sin(2x + 5π/6)
= 1/2 + (1/2)cos2x+ (1/2)cos 2 x– (√3/2)cosxsinx+ 1/2 – (1/2)cos(π/3 – 2x)
= 1 +(1/2)cos2x + (1/2)*(1/2)(1 + cos2x) – (√3/4)sin2x – (1/2)[cos(π/3)cos2x+ sin(π/3)sin2x]
= 5/4 + (3/4)cos2x– (√3/4)sin2x – (1/4)cos2x – (√3/4)sin2x
= 5/4 + (1/2)cos2x – (√3/2)sin2x
= 5/4 + sin(2x + 5π/6)
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- 1楼网友：空山清雨
- 2021-04-29 01:33
降次，拆开就可以了
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