数学题三角函数已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
数学题三角函数已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
数学题三角函数
已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
数学题三角函数已知2sin(3π+θ)=cos(π+θ),求2sin^2 θ+3sinθcosθ-cos^2 θ
2sin(3π+θ)=cos(π+θ)
2sinθ=cosθ
tanθ=1/2
2sin^2 θ+3sinθcosθ-cos^2 θ 2sin^2 θ+3sinθcosθ-cos^2 θ
___________________________= ____________________________=
1 sin^2 θ +cos^2 θ
分子分母同除以cos^2 θ
2tan^2θ+3tanθ-1
________________ 将tanθ=1/2带入
tan^2θ+1
=4/5
答案:
过程:
2sin(3π+θ)=cos(π+θ)
→ 2sin(π+θ)=cos(π+θ)
→ -2sinθ=-cosθ
→ tanθ=sinθ/cosθ=1/2
→ 即θ=π/4 (即45°)
→ 2sin^2θ+3sinθcosθ-cos^2θ
→ =2sin^2θ+(3/2)*sin2θ-(1-sin^2θ)
...
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答案:
过程:
2sin(3π+θ)=cos(π+θ)
→ 2sin(π+θ)=cos(π+θ)
→ -2sinθ=-cosθ
→ tanθ=sinθ/cosθ=1/2
→ 即θ=π/4 (即45°)
→ 2sin^2θ+3sinθcosθ-cos^2θ
→ =2sin^2θ+(3/2)*sin2θ-(1-sin^2θ)
→ =3sin^2θ+(3/2)*sin2θ-1
→ =3*(sin90°)^2+3/2sin90°-1
→ =3*1+3/2-1
→ =1/2
收起
2sin(3π+θ)=cos(π+θ)
2sinθ=cosθ
tanθ=1/2
2sin^2 θ+3sinθcosθ-cos^2 θ 2sin^2 θ+3sinθcosθ-cos^2 θ
___________________________= ____________________________=
1 sin^2 θ +cos^2 θ
2tan^2θ+3tanθ-2
________________=0
tan^2θ+1