若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?

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若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?c

若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?
若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?

若cos(α+π/3)=1/7,α∈(0,π/2),则cosα=?
cos(α+π/3)=1/7,α∈(0,π/2),
则sin(α+π/3)=4√3/7,
cosα=cos[(α+π/3)-π/3]
=cos(α+π/3)cos(π/3)+sin(α+π/3)sin(π/3)
=1/14+12/14=13/14.

∵cos(α+π/3)=1/7,α∈(0,π/2),
∴sin(α+π/3)=4√3/7,
cosα=cos[(α+π/3)-π/3]
=cos(α+π/3)cos(π/3)+sin(α+π/3)sin(π/3)
=1/14+12/14=13/14.

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