積化和差公式
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sinAcosB=(1/2)[sin(A+B)+sin(A-B)] cosAcosB=(1/2)[cos(A+B)+cos(A-B)] sinAsinB=(1/2)[cos(A+B)-cos(A-B)] |
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和差化積公式
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sinx+siny=2sin[(x+y)/2]cos[(x-y)/2] sinx-siny=2cos[(x+y)/2]sin[(x-y)/2] cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2] cosx-cosy=-2sin[(x+y)/2]sin[(x-y)/2] |
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--------------------(18)--------------------
sin(A+B)+sin(A-B)
=[sinAcosB+cosAsinB]+[sinAcosB-cosAsinB]
=2sinAcosB
∴sinAcosB=(1/2)[sin(A+B)+sin(A-B)]
--------------------(19)--------------------
cos(A+B)+cos(A-B)
=[cosAcosB-sinAsinB]+[cosAcosB+sinAsinB]
=2cosAcosB
∴cosAcosB=(1/2)[cos(A+B)+cos(A-B)]
--------------------(20)--------------------
cos(A+B)-cos(A-B)
=[cosAcosB-sinAsinB]-[cosAcosB+sinAsinB]
=2sinAsinB
∴sinAsinB=(1/2)[cos(A+B)-cos(A-B)]
--------------------(21)--------------------
根據(1)，代入A+B=x A-B=y，由此亦得出A=(x+y)/2 B=(x-y)/2
∴sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]
--------------------(22)--------------------
根據(4)，代入y=-y
sinx+sin(-y)=2sin{[x+(-y)]/2}cos{[x-(-y)]/2}
∴sinx-siny=2cos[(x+y)/2]sin[(x-y)/2]
--------------------(23)--------------------
根據(2)，代入A+B=x A-B=y，由此亦得出A=(x+y)/2 B=(x-y)/2
∴cosx+cosy=2cos[(x+y)/2]cos[(x-y)/2]
--------------------(24)--------------------
根據(3)，代入A+B=x A-B=y，由此亦得出A=(x+y)/2 B=(x-y)/2
∴cosx-cosy=-2sin[(x+y)/2]sin[(x-y)/2]