三角函數應用計算

1.
令sinθ = y,則cosθ = 2-3y = √(1-y^2)
由此可得(2-3y)^2 = 1-y^2
10y^2 –12y +3 = 0
y = (6 + √6) /10 [不合,因為會使cosθ小於0] 或 (6 - √6) /10
當sinθ = (6 - √6) /10時, cosθ= (3√6 +2)/10
所以sinθ + cosθ = (4 + √6)/5………….(解答)

2.
tanθ ,cotθ為X^2-3X+1=0之兩根
則tanθ+ cotθ = 1 /(sinθ *cosθ) = 3
sinθ *cosθ = 1/3
(sinθ + cosθ)^2 = 1 + 2 sinθ *cosθ = 5/3
sinθ + cosθ = √15/3……………(解答)

3.
因為角C =90度
所以cosA = sinB = x,sinA = cosB = √(1-x^2)
2x + 3*√(1-x^2) = 3
9(1-x^2) = (3-2x)^2
13x^2 –12x = 0
x = 0(不合)或 12/13

就像流浪的流星所說,應該是sinA/cosA才對
所以原式 = (5/13) /(12/13) = 5/12……………(解答)

2.tanΘ cotΘ為X2-3X+1=0之兩根 且SinΘ>cosΘ>0
求sinΘ+cosΘ 答案 根號15/3
sol:依據根與係數觀係知tanΘ+ cotΘ=3
==>1/(sinΘ*cosΘ ) =3 ==> sinΘ*cosΘ =1/3
sinΘ+cosΘ =根號1+2sinΘ*cosΘ =根號1+2*(1/3)=(根號15)/3

2009-07-09 03:59:59 補充：
1.假設cosΘ+3sinΘ=2 且0<Θ<90 求cosΘ+sinΘ
答 (4+根號6) /5
sol:cosΘ+3sinΘ=2 ==>sinΘ=(2-sinΘ)/3 ==>sin^2Θ=[(2-sinΘ)^2]/9
==>(1-cos^2Θ)=[(2-sinΘ)^2]/9 ==>cosΘ=(2+3*根號6)/10 ,
sinΘ=(6-根號6)/10
所以 cosΘ+sinΘ=(2+3*根號6)/10 +(6-根號6)/10=(4+根號6) /5

第3題應該是sinA/cosA