sin2x +cos2x

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sin2x+cos2xsin2x+cos2xsin2x+cos2x此处采用弧度制asinA+bcosA=√(a^2+b^2)sin(A+M),(tanM=b/a)所以sin2x+cos2x=√(1&s

sin2x +cos2x
sin2x +cos2x

sin2x +cos2x
此处采用弧度制
asinA+bcosA=√(a^2+b^2)sin(A+M) ,(tanM=b/a)
所以sin2x +cos2x=√(1²+1²)sin(2x+M),tanM=1/1=1,
所以M=π/4,
所以sin2x +cos2x=√(1²+1²)sin(2x+M)
=√2sin(2x+π/4)

sin2X+cos2X
=根号2*((根号2)/2*sin2X+(根号2)/2*cos2X)
=根号2*(cos45°*sin2X+sin45°*cos2X)
=根号2*sin(2X+45°)