f(x)=2sin^2(pi/4+x)-√3cos2x化简 谢谢
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f(x)=2sin^2(pi/4+x)-√3cos2x化简谢谢f(x)=2sin^2(pi/4+x)-√3cos2x化简谢谢f(x)=2sin^2(pi/4+x)-√3cos2x化简谢谢f(x)=2s
f(x)=2sin^2(pi/4+x)-√3cos2x化简 谢谢
f(x)=2sin^2(pi/4+x)-√3cos2x
化简 谢谢
f(x)=2sin^2(pi/4+x)-√3cos2x化简 谢谢
f(x)=2sin^2(pi/4+x)-√3cos2x
=[1-cos(2(pi/4+x)]-根号3 cos2x
=1-cos(pi/2+2x)-根号3cos2x
=1+sin2x-根号3cos2x
=1+2[1/2sin2x-根号3 /2 cos2x]
=1+2sin(2x-60)
f(x)=2sin^2(pi/4+x)-√3cos2x
=2[sin(π/4+x)]^2-√3cos2x
=2[(√2/2)*(sinx+cosx)]^2-√3cos2x
=(sinx+cosx)^2-√3cos2x
=1+2sinxcosx-√3cos2x
=1+[sin(2x)-√3cos2x]
=1+2sin(2x-π/3)
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f(x)h(x)=√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)√2/2sin(x+pi/4)cos(x+5pi/4) =-√2/2sin(x+pi/4)cos(x+pi/4)这个是怎么化出来的?
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f(x)=2sin^2(pi/4+x)-√3cos2x化简 谢谢
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