已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
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已知cos(x+π/6)=5/13,x∈(0,π/2),则cosx=
x∈(0,π/2)
x+π/6∈(π/6,2π/3)
所以sin(x+π/6)>0
sin²(x+π/6)+cos²(x+π/6)=1
所以sin(x+π/6)=12/13
cosa
=cos[(x+π/6)-π/6]
=cos(x+π/6)cosπ/6+sin(x+π/6)sinπ/6
=(5√3+12)/26
x+π/6∈(π/6,2π/3)
所以sin(x+π/6)>0
sin²(x+π/6)+cos²(x+π/6)=1
所以sin(x+π/6)=12/13
cosa
=cos[(x+π/6)-π/6]
=cos(x+π/6)cosπ/6+sin(x+π/6)sinπ/6
=(5√3+12)/26
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