1.如果tanA=1又3分2,利用三角恆等式求3sinA-cosA分之4sinA的值
2.如果tanA=n,利用三角恆等式,以n表示cosA-sinA分之4sinA+3cosA的值
數學問題三角比 恆等式要過程
2007-03-02 5:02 am
回答 (3)
2007-03-02 5:29 am
✔ 最佳答案
1.首先，tanA=5/3。(4*sinA) / (3*sinA - cosA)
= (4*tanA) / (3*tanA - 1) (分子分母同時除以 cosA)
= (4*5/3) / (3*5/3 - 1)
= (20/3) / 4
= 5/3
2.
(4*sinA + 3*cosA) / (cosA - sinA)
= (4*tanA + 3) / (1 - tanA) (分子分母同時除以 cosA)
= (4n + 3) / (1 - n)
2007-03-02 5:50 am
1.首先，tanA=5/3。
(4*sinA) / (3*sinA - cosA)
= (4*tanA) / (3*tanA - 1) (分子分母同時除以 cosA)
= (4*5/3) / (3*5/3 - 1)
= (20/3) / 4
= 5/3
2.
(4*sinA + 3*cosA) / (cosA - sinA)
= (4*tanA + 3) / (1 - tanA) (分子分母同時除以 cosA)
= (4n + 3) / (1 - n)
(4*sinA) / (3*sinA - cosA)
= (4*tanA) / (3*tanA - 1) (分子分母同時除以 cosA)
= (4*5/3) / (3*5/3 - 1)
= (20/3) / 4
= 5/3
2.
(4*sinA + 3*cosA) / (cosA - sinA)
= (4*tanA + 3) / (1 - tanA) (分子分母同時除以 cosA)
= (4n + 3) / (1 - n)
2007-03-02 5:39 am
1. tanA=5/3
(4sinA) / (3sinA - cosA) = (4tanA) / (3tanA - 1)
= (20/3) / (5 - 1)
= 5/3
2. tanA = n
(4sinA + 3cosA) / (cosA - sinA) = (4tanA + 3) / (1 - tanA)
= (4n + 3) / (1 - n)
= - 4 + 7 / (1 - n)
(4sinA) / (3sinA - cosA) = (4tanA) / (3tanA - 1)
= (20/3) / (5 - 1)
= 5/3
2. tanA = n
(4sinA + 3cosA) / (cosA - sinA) = (4tanA + 3) / (1 - tanA)
= (4n + 3) / (1 - n)
= - 4 + 7 / (1 - n)
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