求值:sin(π/4+3x)cos(3/π-3x)+cos(π/6+3x)cos(π/4+3x)
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求值:sin(π/4+3x)cos(3/π-3x)+cos(π/6+3x)cos(π/4+3x)
其中:π=派
2.已知sinα=3/5,cosβ=15/17,求cos(α-β)的值
3.已知tanθ=4/3,θ∈(0,π/2),求cos(2π/3-θ)
其中:π=派
2.已知sinα=3/5,cosβ=15/17,求cos(α-β)的值
3.已知tanθ=4/3,θ∈(0,π/2),求cos(2π/3-θ)
1求值:sin(π/4+3x)cos(π/3-3x)+cos(π/6+3x)cos(π/4+3x)
解 ∵(π/3-3x)= π/2- (π/6+3x)
∴ 原式=sin(π/4+3x)sin(π/6+3x)+cos(π/6+3x)cos(π/4+3x)
=cos[(π/4+3x)-(π/6+3x) ] =cos(π/4-π/6)=(√2+√ 6)/4
2、已知sinα=3/5,cosβ=15/17,求cos(α-β)的值
解 ∵( sinα=3/5,cosβ=15/17
∴ cosα=±4/5 sinβ=±8/17
下有四种情况
cos(α-β)=cosαcosβ+sinαsinβ==±84/85或=±36/85
3.tanθ=4/3,θ∈(0,π/2),
所以cosθ=3/5 sinθ=4/5
cos(2π/3-θ)=cos(π-π/3-θ)= - cos(π/3+θ)
=-[1/2cosθ-(√3/2)sinθ]=-3/10+2√3/5
解 ∵(π/3-3x)= π/2- (π/6+3x)
∴ 原式=sin(π/4+3x)sin(π/6+3x)+cos(π/6+3x)cos(π/4+3x)
=cos[(π/4+3x)-(π/6+3x) ] =cos(π/4-π/6)=(√2+√ 6)/4
2、已知sinα=3/5,cosβ=15/17,求cos(α-β)的值
解 ∵( sinα=3/5,cosβ=15/17
∴ cosα=±4/5 sinβ=±8/17
下有四种情况
cos(α-β)=cosαcosβ+sinαsinβ==±84/85或=±36/85
3.tanθ=4/3,θ∈(0,π/2),
所以cosθ=3/5 sinθ=4/5
cos(2π/3-θ)=cos(π-π/3-θ)= - cos(π/3+θ)
=-[1/2cosθ-(√3/2)sinθ]=-3/10+2√3/5
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