3个回答
展开全部
1、原式=sin30°cosα+sinαcos30°=sin(α+30°)
2、原式=2[(√2/2)cosα-(√2/2)sinα=2sin(45°-α)
3、原式=5[(3/5)cosx+(4/5)sinx]=5sin(α+x) (其中α=arcsin(3/5))
4、原式=10[(4/5)sinx-(3/5)cosx]=10sin(α-x) (其中α=arcsin(3/5))
5、原式=√41[(4/√41)cosα+(5/√41)sinα]=√41sin(α+θ) (其中θ=arcsin(4/√41))
6、1+4sin(A/2)sin(B/2)sin(C/2)=1-2sin(C/2)[cos(A+B)/2-cos(A-B)/2]
=1-2sin(c/2)[sin(C/2)-cos(A-B)/2]
=1-2sin^2(C/2)+2sin(C/2)cos[(A-B)/2]
=cosC+2cos[(A+B)/2]cos[(A-B)/2]
=cosC+cosB+cosA
7、sin2A+sin2B+sin2C=sin2A+2sin(B+C)cos(B-C)
=2sinAcosA+2sinAcos(B-C)
=2sinA[cosA+cos(B-C)]
=2sinA[cos(B-C)-cos(B+C)]
=4sinAsinBsinC
2、原式=2[(√2/2)cosα-(√2/2)sinα=2sin(45°-α)
3、原式=5[(3/5)cosx+(4/5)sinx]=5sin(α+x) (其中α=arcsin(3/5))
4、原式=10[(4/5)sinx-(3/5)cosx]=10sin(α-x) (其中α=arcsin(3/5))
5、原式=√41[(4/√41)cosα+(5/√41)sinα]=√41sin(α+θ) (其中θ=arcsin(4/√41))
6、1+4sin(A/2)sin(B/2)sin(C/2)=1-2sin(C/2)[cos(A+B)/2-cos(A-B)/2]
=1-2sin(c/2)[sin(C/2)-cos(A-B)/2]
=1-2sin^2(C/2)+2sin(C/2)cos[(A-B)/2]
=cosC+2cos[(A+B)/2]cos[(A-B)/2]
=cosC+cosB+cosA
7、sin2A+sin2B+sin2C=sin2A+2sin(B+C)cos(B-C)
=2sinAcosA+2sinAcos(B-C)
=2sinA[cosA+cos(B-C)]
=2sinA[cos(B-C)-cos(B+C)]
=4sinAsinBsinC
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询