f(x)=4sin(2∏/3-x)cosx,求f(∏/12)
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f(x)=4sin(2∏/3-x)cosx,求f(∏/12)
f(x)=4sin(2∏/3-x)cosx,求f(∏/12)
f(x)=4sin(2∏/3-x)cosx,求f(∏/12)
f(∏/12)=4sin(2∏/3-∏/12)cos(∏/12)=
4sin(7∏/12)cos(∏/12)=
4sin(∏/3+∏/4)cos(∏/3-∏/4)=
4[sin(∏/3)cos(∏/4)+cos(∏/3)sin(∏/4)][cos(∏/3)cos(∏/4)+sin(∏/3)sin(∏/4)]=
4(根2/4+根6/4)(根6/4+根2/4)=
4(1/8+根3/4+3/8)=2+根3
f(π/12)=4sin(8π/12-π/12)cosπ/12
=4sin(7π/12)cosπ/12
=4sin(π/2+π/12)cosπ/12
=4[sinπ/2*cosπ/12+cosπ/2*sinπ/12]cosπ/12
=4cos²(π/12)
=4{[1+cos(π/6)]/2}
=2+√3
第四行是和差化积公式
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