①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ
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①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ
②证明tanθsinθ∕tanθ-sinθ=1+cosθ∕sinθ③已知sinα-2cosα+1=0,α≠kπ+π∕2,k∈z求:tan(3π-α)和1∕sin2α-sinαcosα+1的值
②证明tanθsinθ∕tanθ-sinθ=1+cosθ∕sinθ③已知sinα-2cosα+1=0,α≠kπ+π∕2,k∈z求:tan(3π-α)和1∕sin2α-sinαcosα+1的值
![①利用公式sin(π-θ)=sinθ和sin(∏+θ)=-sinθ证明:sin(-θ)=-sinθ](/uploads/image/z/15607799-71-9.jpg?t=%E2%91%A0%E5%88%A9%E7%94%A8%E5%85%AC%E5%BC%8Fsin%28%CF%80-%CE%B8%29%3Dsin%CE%B8%E5%92%8Csin%28%E2%88%8F%2B%CE%B8%29%3D-sin%CE%B8%E8%AF%81%E6%98%8E%EF%BC%9Asin%EF%BC%88-%CE%B8%EF%BC%89%3D-sin%CE%B8)
①sin(-θ)=sin[π-(π+θ)]
=sin(π+θ) (∵sin(π-θ)=sinθ)
=-sinθ (∵sin(π+θ)=-sinθ).
②tanθsinθ∕(tanθ-sinθ)=(sinθ)^2/(sinθ-sinθcosθ) (分子分母同乘cosθ)
=[1-(cosθ)^2]/[sinθ(1-cosθ)]
=(1+cosθ)(1-cosθ)/[sinθ(1-cosθ)]
=(1+cosθ)/sinθ.
③∵sinα-2cosα+1=0 ==>sinα=2cosα-1
==>[2cosα-1]^2+(cosα)^2=1
==>5(cosα)^2-4cosα=0
∴cosα(5cosα-4)=0
∵α≠kπ+π∕2,k∈z ==>cosα≠0
∴5cosα-4=0 ==>cosα=4/5
==>sinα=2cosα-1=3/5
故1∕(sin2α-sinαcosα+1)=1/(2sinαcosα-sinαcosα+1) (应用倍角公式)
=1/(sinαcosα+1)
=1/[(3/5)(4/5)+1]
=25/37.
=sin(π+θ) (∵sin(π-θ)=sinθ)
=-sinθ (∵sin(π+θ)=-sinθ).
②tanθsinθ∕(tanθ-sinθ)=(sinθ)^2/(sinθ-sinθcosθ) (分子分母同乘cosθ)
=[1-(cosθ)^2]/[sinθ(1-cosθ)]
=(1+cosθ)(1-cosθ)/[sinθ(1-cosθ)]
=(1+cosθ)/sinθ.
③∵sinα-2cosα+1=0 ==>sinα=2cosα-1
==>[2cosα-1]^2+(cosα)^2=1
==>5(cosα)^2-4cosα=0
∴cosα(5cosα-4)=0
∵α≠kπ+π∕2,k∈z ==>cosα≠0
∴5cosα-4=0 ==>cosα=4/5
==>sinα=2cosα-1=3/5
故1∕(sin2α-sinαcosα+1)=1/(2sinαcosα-sinαcosα+1) (应用倍角公式)
=1/(sinαcosα+1)
=1/[(3/5)(4/5)+1]
=25/37.
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