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高一数学两角和与差的正弦
1、若a为锐角,sin(a-π/6)=1/3,则sina=?
2、已知03.sina-cosb=1/2,cosa-sinb=1/3,则sin(a+b)=?
人气:337 ℃ 时间:2020-09-23 05:44:09
解答
(1)若a为锐角,sin(a-π/6)=1/3, cos(a-π/6)=2根号2/3
sina=sin[(a-π/6)+π/6]=sin(a-π/6)cosπ/6+cos(a-π/6)sinπ/6=(2根号2+3)/6
(2) 已知0cos(a+b)=-4/5,sin(a+b)=-3/5sina=3/5, cosa=4/5.
sinb=sin[(a+b)-a]=sin(a+b)cosa-cos(a+b)sina=0
(3)sina-cosb=1/2,cosa-sinb=1/3,
(sina-cosb)^2=(sina)^2+(cosb)^2-2sina*cosb=1/4
(cosa-sinb)^2=(cosa)^2+(sinb)^2-2cosa*sinb=1/9
相加得到sin(a+b)=59/72
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