y=cos(x-π/6)[√3sin(x-π/6)+cos(x-π/6)]怎么化简(求最值)化简步骤详细一点,

justme8888 1年前 已收到4个回答 举报

醒目的小羊 幼苗

共回答了25个问题采纳率:88% 举报

y=√3sin(x-π/6) cos(x-π/6)+cos(x-π/6)cos(x-π/6)
=√3 /2 sin(2x-π/3)+ 1/2 (cos(2x-π/3)+ 1)
=sin(2x-π/3+π/6)+1/2
=sin (2x-π/6)+1/2
将2x-π/6看成整体,求sin(2x-π/6)的范围,再求y

1年前

5

jingjiniao_009 幼苗

共回答了4个问题 举报

y=√3sin(x-π/6) cos(x-π/6)+cos(x-π/6)cos(x-π/6)
=√3 /2 sin(2x-π/3)+ 1/2 (cos(2x-π/3)+ 1)
=sin(2x-π/3+π/6)+1/2
=sin (2x-π/6)+1/2
将2x-π/6看成整体,求sin(2x-π/6)的范围,再求y

1年前

2

冰水13 幼苗

共回答了3个问题 举报

=2cos(x-π/6)[√3/2sin(x-π/6)+1/2cos(x-π/6)]
=2cos(x-π/6)[sin(x-π/6)+π/6]
=2cos(x-π/6)sinx

1年前

2

kana13 幼苗

共回答了1个问题 举报

原式=2[1/2*c0s(π/6-x)-√3/2*sin(π/6-x)] =2[cos(π/6-x)cosπ/3-sin(π/6-x)sinπ/3] =2cos(π/6-x+π/3) =2cos(π/

1年前

0
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