請問有人會這個嗎？（急）?

1.設兩向量a.b的夾角為θ，且|a|向量=|b|向量，向量|a+b|=4 , 向量|a-b|=3，則cosθ=？

Set { } = 向量, c = cosθ, s = sinθ

Let {a} = (k, 0), {b} = (k*c, k*s) where k = waiting for determined

then {a} + {b} = (k(1+c), k*s)

and {a} - {b} = (k(1-c), -k*s)

|{a} + {b}| = k√[(1+2c+c^2)+s^2] = k√2(1+c) = 4

|{a} - {b}| = k√[(1-2c+c^2)+s^2] = k√2(1-c) = 3

divided:

4/3 = √(1+c)/√(1-c)

=> 16/9 = (1 + c)/(1 - c)

=> 9 + 9c = 16 - 16c

=> 25c = 7

=> c = 7/25 =answer



2.方程式（81x平方）=27分之根號3之解為何？

81x^2 = √3/27

x^2 = √3/(81*27)

x = +- ∜3/(9*3√3)

= +- ∜3*√3/(27*3)

= +- ∜3*∜9 / 81

= +- ∜27 / 81

= answer




3.設向量a=(cos60度, sin30度),向量b=(tan315度, sec120度)，則向量a+b=?

{a} = (1/2, 1/2)

{b} = (-tan45, -sec60) = (-1, -2)

{a} + {b} = (-1/2, -3/2)



4.已知三角形ABC之三邊長為8，9，14，則向量

AB + BC + CA = 0 for a closed loop



5.設 0 <= X < 2π，則函數f(X) = cos²X - 3sinX + 2之最大值為？

We find max at point X = 270 deg by using of graphic method.

max = f(270 deg)

= cos²270 - 3sin270 + 2

= 0 + 3 + 2

= 5

= answer



6.y(x) = tan(x/3) 的週期為？

x/3 = π

=> x = 3π

=> Period = 3π = answer



7.設X為銳角，若sinX + cosX = 7/5，w = tanX + cotX = ?

7/5 = √2(sinX/√2 + cosX/√2)

= √2*sin(x + 45)

= √2*cos(X - 45)

=> sin(X + 45) = 7√2/10 & cos(X - 45) = 7√2/10

=> X = asin(7√2/10) - π/4 & X = acos(7√2/10) + π/4

Add:

2X = asin(7√2/10) + acos(7√2/10) = A + B

sin(2X) = sinAcosB + sinBcosA

= (7√2)^2/100 + [√(1-98/100)]^2

= 98/100 + 2/100

= 1

= sin90

=> 2X = 90

=> X = 45 deg

w = 1 + 1 = 2 = answer



8.若a2分之3平方乘a6分之5平方乘a3分之2平方=ax平方，其中a>0,則?

題意不明.無法演算


9.設 a > 0 且a不等於1，若 log<a>3 + log<a>7 = 3, 則 a = ?

3 = log3/loga + log7/loga

3*log(a) = log(a^3)

= log3 + log7

= log21

=> a^3 = 21

=> a = ∛21 = answer



10.若 log2 = a, log7 = b, 則log35等於下列何者？

(a) 1+a-b (b)1+a-b (c)1-a-b (d) 1-a+b

log35 = log5 + log7

= log(10/2) + log7

= log10 - log2 + log7

= 1 - a + b

= (d)



11.已知 x = 755度，則 x = 第幾象限？

x = 755 - 360*2

= 755- 720

= 35

= IQ


12.f(X) = 3sinX-2的最大值為？

-1 <= sinX <= 1

=> max = f(90 deg)

= 3 - 2

= 1

= answer



13.設A(-1, 4), B(2, 3), 則線段AB之垂直平分方程式為？

(x + 1)^2 + (y - 4)^2 = (x - 2)^2 + (y - 3)^2

2x + 1 - 8y + 16 = -4x + 4 - 6y + 9

answer: 6x - 2y + 4 = 0



14.在座標平面上，若兩平行線 L1: x + 2y = k 與 L2: x + 2y + 4 = 0

距離為根號√20且 k > 0, 則 k = ?

Point A = (0, -2) on L2

distance(A to L1) = |x + 2y - k|/√5

= |0 - 4 - k|/√5

= |k + 4|/√5

= √20

|k + 5| = 10

k + 5 = +-10

k = 10 - 5 = 5 = answer