已知cosα= -3/5 ,α∈(π/2,π),求sin(α+π/3)的值
∵cosα= -3/5 ,α∈(π/2,π)
∴sinα=4/5
sin(α+π/3)
=sinαcos(π/3)+sin(π/3)cosα
=(4/5)*(1/2)+(√3/2)*(-3/5)
=2/5-3√3/10
=(4-3√3)/10
已知sinα=15/17,α∈(π/2,π),求cos(π/3－α)的值
∵sinα= 15/17 ,α∈(π/2,π)
∴cosα=-8/17
cos(π/3-α)
=cosαcos(π/3)+sin(π/3)sinα
=(-8/17)*(1/2)+(√3/2)*(15/17)
=-4/17+15√3/34
=(15√3-8)/34
已知cosα= -5/13,α∈(π,3π/2),求cos(α+π/6)的值
∵cosα= -5/13 ,α∈(π,3π/2)
∴sinα=-12/13
cos(α+π/6)
=cosαcos(π/6)-sin(π/6)sinα
=(-5/13)*(√3/2)+(1/2)*(-12/13)
=-5√3/26-6/13
=(-5√3-12)/26
已知tanα=2,求tan(α－π/4）的值
tan(α-π/4)
=[tanα-tan(π/4)/[1+tanαtan(π/4)]
=(tanα-1)/(1+tanα)
=(2-1)/(1+2)
=1/3
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