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已知平面向量A=(cosa,sina),B=(cosb,sinb),|A-B|=2根号5/5
(1)求cos(a-b)的值(2)0<a<π/2,-π/2<b<0,且sinb=-5/13,求sina的值
人气:197 ℃ 时间:2019-10-11 16:08:41
解答
A-B=(cosa-cosb, sina-sinb)
|A-B|^2=(cosa-cosb)^2+(sina-sinb)^2=2-2(cosacosb+sinasinb)=4/5
cosacosb+sinasinb=3/5
cos(a-b)=cosacosb+sinasinb=3/5
-π/20sin(a-b)=√(1-[sin(a-b)]^2)=4/5
sinb=-5/13,
cosb=√[1-(sinb)^2]=12/13
sina=sin[(a-b)+b]=sin(a-b)cosb+cos(a-b)sinb=33/65
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