已知tanα sinα =α ,tanα -sinα =b,求証:(a方 b方)的平方=16ab 網友廻答:
- 題目有問題,應該是(a²-b²)²=16ab
証:
tanα sinα=a (1)
tanα-sinα=b (2)
[(1) (2)]/2
tanα=(a b)/2
[(1)-(2)]/2
sinα=(a-b)/2
cosα=sinα/tanα=(a-b)/(a b)
sin²α cos²α=1
[(a-b)/2]² [(a-b)/(a b)]²=1
(a-b)²/4 (a-b)²/(a b)²=1
(a-b)²(a b)² 4(a-b)²=4(a b)²
(a²-b²)² 4(a-b)²=4(a b)²
(a²-b²)² 4a²-8ab 4b²=4a² 8ab 4b²
(a²-b²)²=16ab
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