maths高手.............知吾知點計a???

( 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。

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

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