利用和(差)公式化简(1)sinx+√3cosx(2)sinx+cos(3)√3sin2x-cos2x (4)2sin(
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利用和(差)公式化简(1)sinx+√3cosx(2)sinx+cos(3)√3sin2x-cos2x (4)2sin(x+π/3)-2cos(x+π/3)
要过程
要过程
(1)sinx+√3cosx=2(1/2sinx+√3/2cosx)=2(cosπ/3sinx+sinπ/3cosx)=2sin(x+π/3)
(2)sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(cosπ/4sinx+sinπ/4cosx)=√2sin(x+π/4)
(3)√3sin2x-cos2x= 2(√3/2sin2x-1/2cos2x)=2(cosπ/6sin2x-sinπ/6cos2x)=2sin(2x-π/6)
(4)2sin(x+π/3)-2cos(x+π/3) =2√2[√2/2sin(x+π/3)-√2/2cos(x+π/3)]=2√2[cosπ/4sin(x+π/3)-sinπ/4cos(x+π/3)]=2√2sin(x+π/3-π/4)=2√2sin(x+π/12)
(2)sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(cosπ/4sinx+sinπ/4cosx)=√2sin(x+π/4)
(3)√3sin2x-cos2x= 2(√3/2sin2x-1/2cos2x)=2(cosπ/6sin2x-sinπ/6cos2x)=2sin(2x-π/6)
(4)2sin(x+π/3)-2cos(x+π/3) =2√2[√2/2sin(x+π/3)-√2/2cos(x+π/3)]=2√2[cosπ/4sin(x+π/3)-sinπ/4cos(x+π/3)]=2√2sin(x+π/3-π/4)=2√2sin(x+π/12)
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