cos[x+(π/4)]=(√2)/10,x∈(π/2,3π/4),求sinx和sin[2x+(π/3)].
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cos[x+(π/4)]=(√2)/10,x∈(π/2,3π/4),求sinx和sin[2x+(π/3)].
(cosx-sinx)(√2)/2=(√2)/10
cosx-sinx=1/5
cosx=4/5
sinx=3/5
sin2x=24/25
cos2x=7/25
sin[2x+(π/3)]=sin2xcos(π/3)+cos2xsin(π/3)=24/25*1/2+7/25*√3/2
cosx-sinx=1/5
cosx=4/5
sinx=3/5
sin2x=24/25
cos2x=7/25
sin[2x+(π/3)]=sin2xcos(π/3)+cos2xsin(π/3)=24/25*1/2+7/25*√3/2
cos[x+(π/4)]=(√2)/10,x∈(π/2,3π/4),求sinx和sin[2x+(π/3)].
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