tana=1/2,求sin²a-sinacosa+3cos²a已知tan²a=2tan²p+1,求证sin²p=2sin²a-1(这个p是beta,
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![tana=1/2,求sin²a-sinacosa+3cos²a已知tan²a=2tan²p+1,求证sin²p=2sin²a-1(这个p是beta,](/uploads/image/z/10210477-13-7.jpg?t=tana%3D1%2F2%2C%E6%B1%82sin%26%23178%3Ba-sinacosa%2B3cos%26%23178%3Ba%E5%B7%B2%E7%9F%A5tan%26%23178%3Ba%3D2tan%26%23178%3Bp%2B1%2C%E6%B1%82%E8%AF%81sin%26%23178%3Bp%3D2sin%26%23178%3Ba-1%EF%BC%88%E8%BF%99%E4%B8%AAp%E6%98%AFbeta%2C)
tana=1/2,求sin²a-sinacosa+3cos²a已知tan²a=2tan²p+1,求证sin²p=2sin²a-1(这个p是beta,
tana=1/2,求sin²a-sinacosa+3cos²a
已知tan²a=2tan²p+1,求证sin²p=2sin²a-1(这个p是beta,
tana=1/2,求sin²a-sinacosa+3cos²a已知tan²a=2tan²p+1,求证sin²p=2sin²a-1(这个p是beta,
(1)
∵tana=1/2
∴cosa≠0
sin²a-sinacosa+3cos²a
=(sin²a-sinacosa+3cos²a)/(sin²a+cos²a)
=(tan²a-tana+3)/(tan²a+1) (分子分母同时除以cos²a)
=(1/4-1/2+3)/(1/4+1)
=11/4 ÷ 5/4
=11/5
(2)
证明:
tan²a=2tan²β+1
sin²a/cos²a=2sin²β/cos²β+2-1
sin²a/cos²a+1=2sin²β/cos²β+2
(sin²a+cos²a)/cos²a=2(sin²β+cos²β)/cos²β
1/cos²a=2/cos²β
cos²β=2cos²a
1-sin²β = 2(1-sin²a)=2-2sin²a
sin²β=1-(2-2sin²a)=2sin²a-1