1. (sinθ+1)(sinθ-1)+cos二次方θ+2
２ 。 。
2 若cosθ＝—, 其中0 ≦θ≦90 試不用計算機，求sinθ和tanθ的值
3
maths高手.............知吾知點計a???
2007-07-31 3:33 am
回答 (3)
2007-07-31 3:42 am
✔ 最佳答案
( 1 ) (sinθ+1)(sinθ-1)+cos2θ+2= sin2θ- 1 + cos2θ+ 2
= 1 – 1 + 2
= 2
( 2 ) cos θ= 2 / 3
sin θ= √( 32 - 22 ) / 3 = √5 / 3
tan θ= √5 / 2
2007-07-30 19:47:59 補充：
題1 : sin^2 θ + cos^2 θ = 1 題2 : 只要畫一個直角三角形, 鄰邊為2, 斜邊為3, 接著用畢氏定理求出對邊便計到了。
參考: My Maths Knowledge
2007-08-02 5:25 pm
1. (sinθ+1)(sinθ-1)+cos^2 θ+2
= sin^2 θ - 1 + cos^2 θ + 2
=(sin^2 θ + cos^2 θ) + 1
=1 + 1
=2.
2.cosθ = 2/3
sinθ =sqrt (1 - cos^2 θ) <--- By the identity sin^2 θ + cos^2 θ = 1
= sqrt[1 - (2/3)^2]
=sqrt(5) / 3
tanθ = sinθ/cosθ
= [sqrt(5) / 3 ] / (2/3)
= sqrt(5) / 2
= sin^2 θ - 1 + cos^2 θ + 2
=(sin^2 θ + cos^2 θ) + 1
=1 + 1
=2.
2.cosθ = 2/3
sinθ =sqrt (1 - cos^2 θ) <--- By the identity sin^2 θ + cos^2 θ = 1
= sqrt[1 - (2/3)^2]
=sqrt(5) / 3
tanθ = sinθ/cosθ
= [sqrt(5) / 3 ] / (2/3)
= sqrt(5) / 2
2007-07-31 3:43 am
1.(sinθ+1)(sinθ-1)=sinθ^2-1^2
=sinθ^2-1
(sinθ+1)(sinθ-1)+cosθ^2+2=sinθ^2-1+cosθ^2+2
=sinθ^2+cosθ^2-1+2
=1-1+2
=2
2.用畢氏定理
cosθ=2/3=鄰邊/斜邊
對邊就即係開方(3^2-2^2) =開方5
sinθ=開方5/3 tanθ=開方5/2
=sinθ^2-1
(sinθ+1)(sinθ-1)+cosθ^2+2=sinθ^2-1+cosθ^2+2
=sinθ^2+cosθ^2-1+2
=1-1+2
=2
2.用畢氏定理
cosθ=2/3=鄰邊/斜邊
對邊就即係開方(3^2-2^2) =開方5
sinθ=開方5/3 tanθ=開方5/2
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