F.3 Trigonometry

你可以畫番個直角三角形先
１．　sinθ = 1/2, cosθ = √3 /2, tanθ = 1/√3,
(sinθ+cosθ) / (2sinθ+cosθ)
=(1/2+√3 /2)/(2*1/2+√3 /2)
=[(1+√3)/2]/[(2+√3)/2]
=(√3+1) / (2+√3)

2. cosθ = 7/25, sinθ = 24/25 tanθ = 24/7
(sinθ-2cosθ) / (sinθ+3cosθ) * tanθ
=[24/25-2*(7/25)]/[24/25+3*(7/25)]*24/7
=[(24-14)/25]/[(24+21)/25]*24/7
=10/45*24/7
=16/21

1.
sinθ = 1/2
By Pyth. thm,
cosθ= √3
( sinθ+cosθ) / (2sinθ+cosθ)
= ( 1/2 + √3 ) / [ 2(1/2) + √3 ]
= (√3+1) / (2+√3)
2.
cosθ = 7/25
By Pyth. thm,
sinθ = 24/25
tanθ = 24/7
(sinθ-2cosθ) / (sinθ+3cosθ) * tanθ
= [ 24/25- 2( 7/25) ] / [ (24/25)+ 3(7/25) ]* (24/7)
= (2/5) / (9/5) * (24/7)
= 16/21

1.你可以畫一個直角三角形,最長的邊為2.垂直的邊為1. 再用畢氏定理找出第三條邊.
設對角線為1的角be θ , 再找出這角的 sinθ, cosθ. 再代下去 (sinθ+cosθ) / (2sinθ+cosθ).
便可找出答案.
2.畫一個直角三角形,不過這次最長的邊為25.底邊為7. 再用畢氏定理找出第三條邊.
let angle between 25 and 7 be θ 再找出這角的 sinθ, cosθ.和tanθ 再代下去(sinθ-2cosθ) / (sinθ+3cosθ) * tanθ .