sin2x-cos2x-1=√2 sin(2x-π/4) 这又是怎么化的?
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![sin2x-cos2x-1=√2 sin(2x-π/4) 这又是怎么化的?](/uploads/image/z/3859178-50-8.jpg?t=sin2x-cos2x-1%3D%E2%88%9A2+sin%282x-%CF%80%2F4%29+%E8%BF%99%E5%8F%88%E6%98%AF%E6%80%8E%E4%B9%88%E5%8C%96%E7%9A%84%3F)
sin2x-cos2x-1=√2 sin(2x-π/4) 这又是怎么化的?
sin2x-cos2x-1=√2 sin(2x-π/4) 这又是怎么化的?
sin2x-cos2x-1=√2 sin(2x-π/4) 这又是怎么化的?
sin2x-cos2x-1
=√2(sin2x×√2/2-cos2x×√2/2)-1
=√2sin(2x-π/4)-1
如果本题有什么不明白可以追问,
sin2x-cos2x
=√2( sin2x cos(π/4) -cos2xsin(π/4) )
=√2sin(2x-π/4)
sin(2x-π/4)
=sin2xcosπ/4-cos2xsinπ/4
=sin2x-cos2x-1