设α为锐角,若cos(α+π/6)=4/5,则sin(2α+π/6)=

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 10:08:31
设α为锐角,若cos(α+π/6)=4/5,则sin(2α+π/6)=
xSJ@ mM:$1I~D\+7m?PHA])J@!SU;ɫN2sιĨ['U.x AgxVk]{}5ڽ6U ľoW)8@K5|fD^]r5](H '?iL~Tg/rgN1*٢ 54uЄ^9ʪEJEn٩@ܶ1eRuX>NX!v۟v05(!Sb!4բ^BdFjex]l( ӯi=JnϒBس6t@FMqV6FݑI9U.g`h>A$5"\T@Z8}6EؘM\R3o^H)1{jꡗ= MD.ל-{84|B

设α为锐角,若cos(α+π/6)=4/5,则sin(2α+π/6)=
设α为锐角,若cos(α+π/6)=4/5,则sin(2α+π/6)=

设α为锐角,若cos(α+π/6)=4/5,则sin(2α+π/6)=
由cos(α+π/6)=4/5 推导出→sin(α+π/6)=3/5
引入一个sin(2α+π/3)
=sin2(α+π/6)
=2sin(α+π/6)cos(α+π/6)
=2×3/5×4/5
=24/25 推导出→cos (2α+π/3)=7/25
sin(2α+π/6)=sin[(2α+π/3)-π/6]
=sin (2α+π/3)cos π/6-cos (2α+π/3)sinπ/6
=24/25×(根号3分之2)- 7/25×2分之一
=(24根号3-7)/50

α∈(0,π/2)
α+π/6∈(π/6,2π/3)
cos(α+π/6)=4/5>0
∴α+π/6∈(π/6,π/2)
∴2α+π/3∈(π/3,π)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(π/3,π/2...

全部展开

α∈(0,π/2)
α+π/6∈(π/6,2π/3)
cos(α+π/6)=4/5>0
∴α+π/6∈(π/6,π/2)
∴2α+π/3∈(π/3,π)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(π/3,π/2)
sin(2α+π/3)=24/25
sin(2α+π/12)
=sin(2α+π/3-π/4)
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*√2/2-7/25*√2/2
=24√2/50-7√2/50
=17√2/50

收起