设α属于(-π,0),且cos(α+π/6)=4/5,则sin(2α+π/12)的值为
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设α属于(-π,0),且cos(α+π/6)=4/5,则sin(2α+π/12)的值为
α∈(-π,0)
α+π/6∈(-5π/6,π/6)
cos(α+π/6)=4/5>0
∴α+π/6∈(-π/6,π/6)
∴2α+π/3∈(-π/3,π/3)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(-π/3,π/3)
sin(2α+π/3)=+-24/25
sin(2α+π/12)
=sin(2α+π/3-π/4)
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*√2/2-7/25*√2/2或=-24/25*√2/2-7/25*√2/2
=24√2/50-7√2/50或=-24√2/50-7√2/50
=17√2/50或-31√2/50
α+π/6∈(-5π/6,π/6)
cos(α+π/6)=4/5>0
∴α+π/6∈(-π/6,π/6)
∴2α+π/3∈(-π/3,π/3)
cos(2α+π/3)
=2cos²(α+π/6)-1
=2*(4/5)²-1
=32/25-1
=7/25>0
∴2α+π/3∈(-π/3,π/3)
sin(2α+π/3)=+-24/25
sin(2α+π/12)
=sin(2α+π/3-π/4)
=sin(2α+π/3)cosπ/4-cos(2α+π/3)sinπ/4
=24/25*√2/2-7/25*√2/2或=-24/25*√2/2-7/25*√2/2
=24√2/50-7√2/50或=-24√2/50-7√2/50
=17√2/50或-31√2/50
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