求证:cosx+sinx=根号2cos(x-π/4)

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求证:cosx+sinx=根号2cos(x-π/4)求证:cosx+sinx=根号2cos(x-π/4)求证:cosx+sinx=根号2cos(x-π/4)cosx+sinx=√2[((√2/2)co

求证:cosx+sinx=根号2cos(x-π/4)
求证:cosx+sinx=根号2cos(x-π/4)

求证:cosx+sinx=根号2cos(x-π/4)
cosx+sinx
=√2[((√2/2)cosx+(√2/2)sinx]
=√2[cosxcos(π/4)+sinxsin(π/4)]
=√2cos(x-π/4)
得证

展开即得

两边同时平方 然后右侧化消次就行了