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已知sin(a+2b)=4/5,cos(a+b)=12/13(a+2b,a+b)属于(0~2π)求sinb
人气:453 ℃ 时间:2020-03-30 07:19:34
解答
(a+2b,a+b)属于(0~2π)
sin(a+2b)=4/5,cos(a+2b)=±根号(1-sin^2(a+2b))=±3/5
cos(a+b)=12/13 ,sin(a+b)=±根号(1-cos^2(a+b))=±5/13
sin(a+2b)cos(a+b)=4/5 * 12/13 = 48/65
cos(a+2b)sin(a+b)=(±3/5) * (±5/13)=15/65,或-15/65
sinb=sin{(a+2b)-(a+b)}
=sin(a+2b)cos(a+b)-cos(a+2b)sin(a+b)
=48/65-15/65=33/65
或:
=48/65-(-15/65)=63/65
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