(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).(2)已知cos[(π/4)-a]=3/5,sin[(3π/4)+b]=5/13,其中π/4
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![(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).(2)已知cos[(π/4)-a]=3/5,sin[(3π/4)+b]=5/13,其中π/4](/uploads/image/z/5383002-66-2.jpg?t=%281%29%E5%B7%B2%E7%9F%A5%3Asina%2Bcosb%3D3%2F4%2Ccosa%2Bsinb%3D-5%2F4%2C%E6%B1%82sin%28a%2Bb%29.%282%29%E5%B7%B2%E7%9F%A5cos%5B%28%CF%80%2F4%29-a%5D%3D3%2F5%2Csin%5B%283%CF%80%2F4%29%2Bb%5D%3D5%2F13%2C%E5%85%B6%E4%B8%AD%CF%80%2F4)
(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).(2)已知cos[(π/4)-a]=3/5,sin[(3π/4)+b]=5/13,其中π/4
(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).
(2)已知cos[(π/4)-a]=3/5,sin[(3π/4)+b]=5/13,其中π/4
(1)已知:sina+cosb=3/4,cosa+sinb=-5/4,求sin(a+b).(2)已知cos[(π/4)-a]=3/5,sin[(3π/4)+b]=5/13,其中π/4
1\ Sina+cosb=3/4 cosa+sinb=5/4
(SinA)2+(cosB)2+2sinAcosB=9/16 , (cosA)2+(sinB)2+2cosAsinB=25/16
两式相加 2+2(sinAcosB+cosAsinB)=34/16
Sin(A+B)=1/16
(2)cos[(π/4)-a]=3/5=根号2/2(cosa+sina),①
sin[(3π/4)+b]=5/13=根号2/2(cosb-sinb)②
由此求出(cosa+sina)和(cosb-sinb)的值,再有上面那位写的解第一题地方法可算出cos(a+b)的值,因为π/4
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(2)cos[(π/4)-a]=3/5=根号2/2(cosa+sina),①
sin[(3π/4)+b]=5/13=根号2/2(cosb-sinb)②
由此求出(cosa+sina)和(cosb-sinb)的值,再有上面那位写的解第一题地方法可算出cos(a+b)的值,因为π/4
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