三角比和三角的應用(sin,cos,tan) 10分

1(a)
sin30°tan60°
=1/2 × √3
=√3/2

(b)
cos60°tan60°/sin60°
=cos60°sin60°÷cos60°÷sin60°
=1

(c)
sin90°/cos² 45°
=1/(1÷√2)²
=1/(1÷2)
=2

(d)
tan60°sin60°-tan45°
=√3×√3/2 - 1
=3/2 - 1
=1/2

2(a)
tan²θcos²θ+cos²θ
=sin²θ÷cos²θ×cos²θ + cos²θ
=sin²θ + cos²θ
=1

(b)
cos(90°-θ)/cosθ - 1/tan(90°-θ)
=sin θ/cos θ - 1/(1÷tan θ)
=tan θ - tan θ
=0

3
(a)
sin²θ + cos²θ = 1
sin²θ + (5/13)² = 1
sin²θ = 144/169
sin θ = 12/13 或 -12/13

b)
tan θ
=sin θ÷cos θ
=12/13÷5/13 或 -12/13÷5/13
=12/5或-12/5

(c)
sinθ/tanθ-cosθ
=sin θ÷sin θ/cos θ - cos θ
=cos θ - cos θ
=0

4(a)
√3cosθ=sinθ
tan θ = √3
θ = 180n°+ 60°, n=0,±1,±2,... ...

(b)
2sin2θ=1
sin 2θ = 1/2
2θ = 180n°+ (-1)^n (30°)
θ = 90n°+ (-1)^n (15°)

(c)
sin(3θ+25°) = cos(35°-θ)
sin(3θ+25°) - cos(35°-θ) = 0
sin(3θ+25°) - sin(55°+ θ) = 0
2cos(2θ + 40°)sin(θ - 15°) = 0
cos(2θ + 40°) = 0 或 sin(θ - 15°) = 0
2θ + 40°= 360n°± 90°或 θ - 15°= 180n°
θ = 180n°+ 25°或 180n° - 65°或 180n°+ 15°

(d)
√3 sin^2 45°-sin(90°-θ)=0
√3/2 - cos θ = 0
cos θ = √3/2
θ = 360n°± 30°

2007-08-23 16:05:33 補充：
第4題的a,b,c,d在第一行出現"n"的後面也要寫上"n=0,±1,±2,... ...",以表示n只可以是整數

2007-08-24 12:47:54 補充：
留意第3題cosθ =5/13這是題目給予的但並沒有說明在哪一個象限因為cosθ＞0,可以在象限1,也可以在象限4因此sinθ和tanθ也是有可能是正數或負數敬請留意

1a)Ans: sin30"tan60" = (1/2)(開方3)
= 開方3/2
b)Ans: cos60"tan60°/sin60 = (1/2)(開方3)/(開方3/2)
= (開方3/2)/(開方3/2)
= 1
c)Ans: sin90°/cos^2 45°= 1/( 開方2/2)^2
= 1/(1/2)
= 2

d)Ans: tan60°sin60°-tan45 = (開方3)(開方3/2) - 1
= 3/2 - 1
　= 1/2

2(a)Ans: tan^2θcos^2θ+cos^2θ
= (sin^2θ/cos^2θ)(cos^2θ)+cos^2θ
= sin^2θ+cos^2θ
= 1

(b)Ans: cos(90°-θ)/cosθ - 1/tan(90°-θ)
= sinθ/cosθ - 1/(1/tanθ)
= tanθ - tanθ
= 0

3 若cosθ =5/13,
(a)Ans: sinθ = 開方(1-cos^2θ)
= 開方(1-(5/13)^2)
= 開方(1-25/169)
= 開方(144/169)
= 12/13

(b)Ans: tanθ = sinθ/cosθ
= (12/13)/(5/13)
= 12/5

(c)Ans: sinθ/tanθ-cosθ = (12/13)/(12/5) - (5/13)
= 5/13 - 5/13
= 0

4a)Ans: 開方3cosθ=sinθ
　　　　　　 開方3=sinθ/cosθ　　
tanθ=開方3
θ=60" 　　

b)Ans: 2sin2θ=1
sin2θ=1/2
2θ=30"
θ=15"

c)Ans: sin(3θ+25°) = cos(35°-θ)
cos(90"-(3θ+25")) = cos(35"-θ)
"90-3θ-25" = 35"-θ
90"-25"-35" = -θ+3θ
2θ = 30"
θ = 15"

d)Ans: 開方3 sin^2 45°-sin(90°-θ) = 0
　　　　　　　　　　　　 開方3(開方2/2)^2 = cosθ
cosθ = 開方3(1/2)　
cosθ = 開方3/2
θ = 30"　　　

註：第四題不知道你答案是那一個介限,我是以 0"<= θ <= 90"作準,另外你要記得以下的方程式　　　　 sin 0" = 0 cos 0 = 1 tan 0" = 0
sin 30" = 1/2 cos 30" = 開方3/2 tan 30" = 1/開方3
sin 60" = 開方3/2 cos 60" = 1/2 tan 60" = 開方3
sin 90" = 1 cos 90" = 0 tan 90" = 無限大

　　　　　　sin^2θ+cos^2θ = 1
sinθ/cosθ = tanθ
cosθ = sin(90"-θ)
sinθ = cos(90"-θ)

附加資料：如果你讀到中四理科的A.MATH,會有幾個符號,你有興趣才看啦:

cscθ = 1/sinθ
secθ = 1/cosθ
cotθ = 1/tanθ
1+cot^2θ = csc^2θ
1+tan^2θ = sec^2θ