cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
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cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4)
利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.
已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
cos(x-π/4)+ sin(x-π/4) cos(x-π/4)- sin(x-π/4) 利用什么加法公式(不知道是神马公式),把cos(x-π/4)和 sin(x-π/4)带入下列公式中,求数.已知 1)cos x + sin x = 根号2,2)cos x - sin x =根号2
cos x + sin x = 根号2(sinπ/4cosx+cosπ/4sinx)=(根号2 )sin(π/4+x)
cos x - sin x = 根号2(sinπ/4cosx-cosπ/4sinx)=(根号2 )sin(π/4-x)
直接用和差的正弦(余弦)公式即可
cos(x-π/4)+ sin(x-π/4)
=2sqrt(2)sinx
cos(x-π/4)- sin(x-π/4)
=2sqrt(2)cosx
已知 1)cos x + sin x = 根号2, 2)cos x - sin x =根号2
两式相加(减),得
cosx=sinx=sqrt(2)/2
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直接用和差的正弦(余弦)公式即可
cos(x-π/4)+ sin(x-π/4)
=2sqrt(2)sinx
cos(x-π/4)- sin(x-π/4)
=2sqrt(2)cosx
已知 1)cos x + sin x = 根号2, 2)cos x - sin x =根号2
两式相加(减),得
cosx=sinx=sqrt(2)/2
所以x=π/4+2kπ,k是整数。
【sqrt表示根号】
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根据cosx+sinx=√2,√2*(√2/2*cosx+√2/2*sinx)=√2即√2*(cos(π/4)*cosx+sin(π/4)*sinx)=√2即cos(π/4)*cosx+sin(π/4)*sinx=1,所以cos(x-π/4)=1。.同理根据cos x - sin x =√2,,√2*(√2/2*cosx-√2/2*sinx)=√2即√2*(sin(π/4)*cosx-cos(π/...
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根据cosx+sinx=√2,√2*(√2/2*cosx+√2/2*sinx)=√2即√2*(cos(π/4)*cosx+sin(π/4)*sinx)=√2即cos(π/4)*cosx+sin(π/4)*sinx=1,所以cos(x-π/4)=1。.同理根据cos x - sin x =√2,,√2*(√2/2*cosx-√2/2*sinx)=√2即√2*(sin(π/4)*cosx-cos(π/4)*sinx)=√2即sin(π/4)*cosx-cos(π/4)*sinx=1,所以sin(x-π/4)=-1。所以cos(x-π/4)+ sin(x-π/4) =0,cos(x-π/4)- sin(x-π/4) =2。.
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cosx+sinx=√2 ,cosx-sinx=√2
cosx=√2,sinx=0
cos(x-π/4)+ sin(x-π/4)
=cosxcosπ/4+sinxsinπ/4 +sinxcosπ/4-cosxsinπ/4
=2sinxcosπ/4=0
cos(x-π/4)- sin(x-π/4)
=cosxcosπ/4+sinxsinπ/4- (sinxcosπ/4-cosxsinπ/4)
=2cosxsinπ/4=2