若a∈(0,π/2),且cos(a+π/6)= (—根号2) / 4,则cosa=
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若a∈(0,π/2),且cos(a+π/6)=(—根号2)/4,则cosa=若a∈(0,π/2),且cos(a+π/6)=(—根号2)/4,则cosa=若a∈(0,π/2),且cos(a+π/6)=(
若a∈(0,π/2),且cos(a+π/6)= (—根号2) / 4,则cosa=
若a∈(0,π/2),且cos(a+π/6)= (—根号2) / 4,则cosa=
若a∈(0,π/2),且cos(a+π/6)= (—根号2) / 4,则cosa=
因为:a∈(0,π/2),所以:a+π/6∈(π/6,2π/3),所以:sin(a+π/6)= =根号14/4
所以:cosa=cos[(a+π/6)-π/6]=cos(a+π/6)cosπ/6+sin(a+π/6)sinπ/6
=-根号2/4*根号3/2+根号14/4*1/2
=(根号14-根号6)/8
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若a∈(0,π/2),且cos(a+π/6)= (—根号2) / 4,则cosa=
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