已知tan^a=2tan^b+1 求证 sin^b=2sin^a-1

已知tan^a=2tan^b+1 求证 sin^b=2sin^a-1

tan^a = 2tan^b+1
1+tan^a = 2tan^b+2
sec^a = 2sec^b
cos^b = 2cos^a
1-sin^b = 2-2sin^a
sin^b = 2sin^a-1

sin²a =sin²a/1 =sin²a/(sin²a+cos²a) =(sin²a/cos²a)/[(sin²a/cos²a)+(cos²a/cos²a)] =tan²a/(tan²a+1) 即：sin²a=tan²a/(tan²a+1) 同理：sin²b=tan²b/(tan²b+1) 而：tan²a=2tan²b+1 ∴sin²a =tan²a/(tan²a+1) =(2tan²b+1)/(2tan²b+1+1) =(2tan²b+1)/2(tan²b+1) 则：2sin²a=(2tan²b+1)/(tan²b+1) 那么：2sin²a-1      =(2tan²b+1)/(tan²b+1)-1      =(2tan²b+1-tan²b-1)/(tan²b+1)      =tan²b/(tan²b+1)      =sin²b 即：sin²b=2sin²a-1

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