高中三角函數四題題目

1.
所用公式:
sin2A + cos2A = 1
cosA/sinA = cotA

f(A)
= (2 + cos2A)/(sin2A + sinAcosA)
= [2(sin2A + cos2A) + cos2A]/(sin2A + sinAcosA)
= (2sin2A + 3cos2A)/(sin2A + sinAcosA)
= [(2sin2A + 3cos2A)/sin2A]/[(sin2A + sinAcosA)/sin2A]
= (2 + 3cot2A)/(1 + cotA)
= [2 + 3(-3)2]/[1 + (-3)]
= 29/(-2)
= -29/2


2.
所用公式:
y2 + 4y + 4 = (y + 2)2，即sin2θ + 4sinθ + 4 = (sinθ + 2)2
sin2θ+ cos2θ= 1， 即 cos2θ= 1 - sin2θ

y = (1/4)cos2θ - sinθ
y = (1/4)(1 - sin2θ) - sinθ
y = (1/4) - (1/4)sin2θ - sinθ
y = (1/4) - 1/4(sin2θ + 4sinθ + 4) + (1/4)(4)
y = (5/4) - 1/4(sinθ + 2)2

-1 ≤ sinθ ≤ 1

當 sinθ = -1時, (sinθ + 2)2 的最小值 = (-1 + 2)2 = 1
y 的最大值 = (5/4) - (1/4)(1)
y 的最大值 = 1

當 sinθ = 1 時, (sinθ + 2)2 的最大值 = (1 + 2)2 = 9
y 的最小值 = (5/4) - (1/4)(9)
y 的最小值 = -1


3.
公式：sin(π - A) = sinA

公式：cosAsinB = [sin(A + B) - sin(A - B)]/2
所以：
cos2x.sinx = (sin3x - sinx)/2
cos4x.sinx = (sin5x - sin3x)/2
cos6x.sinx = (sin7x - sin5x)/2
cos8x.sinx = (sin9x - sin7x)/2
cos10x.sinx = (sin11x - sin9x)/2

設 x = π/11

左方
= cos(2π/11) + cos(4π/11) + cos(6π/11) + cos(8π/11) + cos(10π/11)
= cos2x + cos4x + cos6x + cos8x + cos10x
= (cos2x + cos4x + cos6x + cos8x + cos10x).(sinx/sinx)
= (cos2x.sinx + cos4x.sinx + cos6x.sinx + cos8x.sinx + cos10x.sinx)/sinx
= [(sin3x-sinx)/2 + (sin5x-sin3x)/2 + (sin7x-sin5x)/2 + (sin9x-sin7x)/2 + (sin11x-sin9x)/2]/sinx
= (sin3x - sinx + sin5x - sin3x + sin7x - sin5x + sin9x - sin7x + sin11x - sin9x)/(2sinx)
= (sin11x - sinx)/(2sinx)
= [sin11(π/11) - sinx]/(2sin)
= (sinπ - sinx)/(2sinx)
= (0 - sinx)/(2sinx)
= -sinx/(2sinx)
= -1/2
= 右方


4.
所用公式：
tanA = sinA/cosA
sin2A + cos2A = 1 即 cos2A = 1- sin2A
cos2A = 2cos2A - 1

已知： tan2α = 2tan2β + 1
sin2α/cos2α = 2sin2β/cos2β + 1
sin2α/cos2α + 1 = 2sin2β/cos2β + 2
(sin2α + cos2α)/cos2α = (2sin2β + 2cos2β)/cos2β
1/cos2α = 2(sin2β + cos2β)/cos2β
1/cos2α = 2/cos2β
交叉相乘得： 2cos2α = cos2β
2cos2α - cos2β = 0
2cos2α - (1 - sin2β) = 0
(2cos2α - 1) + sin2β = 0
cos2α + sin2β = 0

1. 已知 cotA= －3
f(A) = (2+cos^2A) / (sin^2A + sinAcosA) = ?
Sol
cotA=cosA/sinA=－3 =>cosA=－3sinA，cos^2A=9/10，sin^2 A=1/10
(2+cos^2A) / (sin^2A + sinAcosA)
= (2+cos^2A) / (sin^2A－3 sin^2A)
=(2+9/10)/(－2*1/10)
=(20+9)/(－2)
=－29/2

2. 求 y = 1/4 cos^2θ－sinθ 的最大值和最小值
4y= cos^2θ－4sinθ
=1－sin^2θ－4sinθ
=－(sin^2θ+4sinθ)+1
=－(sin^2θ+4sinθ+4)+5
=－(sinθ+2)^2+5
－1<= sinθ<=1
1<= sinθ+2<=3
1<=(sinθ+1)^2<=9
－9<=－(sinθ+2)^2<=－1
－4<=－(sinθ+1)^2+5<=4
－1<=y<=1

3. 請證明 cos(2pi/11)+cos(4pi/11)+cos(6pi/11)+cos(8pi/11)+cos(10pi/11)=－1/2
Sol
A= cos(2pi/11)+cos(4pi/11)+cos(6pi/11)+cos(8pi/11)+cos(10pi/11)
2Asin(2pi/11)
= 2sin(2pi/11)cos(2pi/11)+ 2sin(2pi/11)cos(4pi/11)+ 2sin(2pi/11)cos(6pi/11)
+ 2sin(2pi/11)cos(8pi/11)+ 2sin(2pi/11)cos(10pi/11)
= sin(4pi/11)+sin(6pi/11)－sin(2pi/11)+sin(8pi/11)－sin(4pi/11)+sin(10pi/11)
－sin(6pi/11)+sin(12pi/11)－sin(8pi/11)
=－sin(2pi/11)
A=－1/2

4. 已知 tan^2α= 2tan^2β + 1，請證明cos2α + sin^2β = 0
Sol
tan^2α=2tan^2β + 1
tan^2α+1=2tan^2β + 2
sec^2α= 2sec^2β
1/cos^2α= 2/cos^2 β
2cos^2α= cos^2β
2cos^2α=1－sin^2β
2cos^2α－1=－sin^2β
cos2α=1－sin^2β
cos2α + sin^2β=0

3. 同乘sin (2π/11)，再用二倍角公式及積化和差公式化簡可得證
4. 先半角公式sin^2β=(1-cos2β)/2
　再用萬能公式cos2α=(1-tan^2α)/(1+tan^2α)代入化簡可得證

2009-07-26 14:53:02 補充：
1. f(A)原式上下同除以sin^2A即可化為cotA的表示法，可不分兩情形討論

1.
tanA=-1/3 < 0 => A是第二或第四象限角，對邊:鄰邊=1:3
=> 斜邊=√(1^2+3^2)=√10

(1)若A是第二象限角：
sinA=+對邊/斜邊=+1/√10，cosA=-鄰邊/斜邊=-3/√10
f(A)= (2+9/10) / (1/10-3/10) = 29/(-2) =－29/2

(2)若A是第四象限角：
sinA=-對邊/斜邊=-1/√10，cosA=+鄰邊/斜邊=+3/√10
f(A)= (2+9/10) / (1/10-3/10) = 29/(-2) =－29/2
∴Ans:－29/2

2009-07-26 11:37:05 補充：
2.
令cosθ=c，sinθ=s
因為s^2+c^2=1 => c^2=1-s^2
代入y=(1/4)(1-s^2) - s = (-1/4)(s^2+4s-1) = (-1/4)[(s+2)^2-5)

因為 -1 <= s <= 1 => 1 <= s+2 <= 3 => 1 <= (s+2)^2 <= 9
=> -4 <= (s+2)^2-5 <= 4
=> 1 >= (-1/4)[(s+2)^2-5) >= -1
∴Ans: 最大值=1，最小值=-1

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