急! 急! 明天考 F5 有關Solve數1條唔識 ?

1.
(a)
cosA + cosB = (2/3)k ...... [1]
cosAcosB = -(1/3)k² ...... [2]

[1]² - [2]*4 :
(cos²A + 2cosAcosB + cos²B) - 4cosAcosB = (16/9)k²
(cosA - cosB)² = (16/9)k²
cosA - cosB = ±(4/3)k ...... [3]

[1] + [3] :
2cosA = 2k or cosA = -(2/3)k
cosA = k or cosA = -k/3

Put cosA = k into [2] :
kcosB = -(1/3)k²
cosB = -k/3

Put cosB = -k/3 into [2] :
(-k/3)cosB = -(1/3)k²
cosB = k

Hence, cosA = k and cosB = -k/3
or cosA = -k/3 and cosB = k


*****
(b)
For all real value of A and B :
-1 ≤ cosA ≤ 1 and -1 ≤ cosB ≤ 1
-1 ≤ k ≤ 1 and -1 ≤ -k/3 ≤ 1
-1 ≤ k ≤ 1 and -1 ≤ k/3 ≤ 1
-1 ≤ k ≤ 1 and -3 ≤ k/3 ≤ 3

Hence, range of values of k :
-1 ≤ k ≤ 1


*****
(c)(i)
Since ∠A is in Quadrant I and ∠B is in Quadrant II,
then cosA = k and cosB = -k/3

cosB = -k/3
cos(90 + A) = -k/3
As A in Quadrant I, sinA = k/3

tanA = sinA/cosA
tanA = (k/3) / k
tanA = 1/3
∠A = 18.43°

cosA = k
cos18.43° = k
k = 0.9487