Question

已知tan^2阿尔法=2tan^2贝塔+1,求证：sin^2贝塔=2sin^2阿尔法-1

Answer 1

tan²α=sin²α/cos²α=sin²α/(1-sin²α)
tan²β=sin²β/coβ²α=sin²β/(1-sin²β)
代入第一个等式，得
sin²α/(1-sin²α)=2sin²β/(1-sin²β)+1
通分化简即得要证的结果
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Answer 2

(sinα/cosα)^2=2(sinβ/cosβ)^2+1 两边乘以(cosαcosβ)^2，移项化简
(cosβ)^2((sinα)^2-(cosα)^2)=2(sinβcosα)^2
(1-(sinβ)^2)(2(sinα)^2-1)=2(sinβ)^2(1-(sinα)^2)展开，化简
2(sinα)^2-1=(sinβ)^2

Answer 3

san阿尔法的平方+san贝塔的平方=1，

Answer 4

sin²α/cos²α=2sin²β/cos²β+1
sin²α/(1-sin²α)=(sin²β+1)/(1-sin²β)
sin²α(1-sin²β)=(1-sin²α)(sin²β+1)
sin²α-sin²βsin²α=sin²β-sin²αsin²β+1-sin²α
sin²β=2sin²α-1

Answer 5

原式=sin^a/cos^a=2sin^b/cos^b+1
sin^acos^b-sin^bcos^a=sin^bcos^a+cos^acos^b=cos^a=1-sin^a
sin^a(1-sin^b)-sin^bcos^a=1-sin^a
sin^a-sin^b(sin^a+cos^a)=1-sin^a
sin^a-sin^b=1-sin^a
2sin^2a-1=sin^b