A maths 兩題 10 分

dom123

2008-03-26 12:43 am

1a)If cosθ-√3 sinθ=rcos(θ+α), where r>0 and 0≦α≦ pi/2
find r andα
b)Hence find the maximum value of √(3-2cos²θ-√3sin2θ) and
the corresponding angleθ,where 0 ≦θ≦ 2pi

2 If A+B+C= 180 , prove that sin²A+sin²B+sin²C-2 = 2cosAcosBcosC

回答 (1)

Audrey Hepburn

2008-03-26 3:12 am

✔ 最佳答案

As follows~~~
As follows~~~
1.a., r > 0 and 0 <= α <= π/2
rcos(θ + α)
= r[cosθcosα - sinθsinα]
= (rcosα)cosθ – (rsinα)sinθ
Comparing coefficients,
rcosα = 1 ───(1)
rsinα = √3─── (2)
(2)/(1): tanα = √3
α = π/3
r = 2

b. √(3 – 2cos2θ - √3 sin2θ)
= √[3 – 2cos2θ - 2√3 sinθcosθ]
= √[3 – 2cosθ(cosθ - √3 sinθ)]
= √[3 – 4cosθcos(θ + π/3)]
= √{3 – 2[cos(2θ + π/3) -cos(π/3)]}
= √[4 – 2cos(2θ + π/3)]
The maximum occurs when cos(2θ + π/3) =-1
And the maximum value = √6
cos(2θ + π/3) = -1
2θ +π/3 = π, 3π
θ = π/3 or 4π/3

2. Prove sin2A + sin2B + sin2C– 2 = 2cosAcosBcosC, for A + B + C = 180˚

R.H.S. = 2cosAcosBcosC
= 2{1/2 [cos(A + B) + cos(A - B)]}cosC
= cos(A + B)cosC + cos(A - B)cosC
= cos(180˚ - C)cosC + cos(A - B)cos[180˚ - (A + B)]
= -cos2C - cos(A - B)cos(A + B)
= -(1 – sin2C) – 1/2 {cos[(A - B) + (A + B)] +cos[(A - B) – (A + B)]}
= - 1 + sin2C – 1/2 [cos2A + cos(-2B)]
= -1 + sin2C – 1/2 (1 – 2sin2A) –1/2 (1 – 2sin2B)
= sin2A + sin2B + sin2C –2
= L.H.S.

參考: Myself~~~

👍 0 👎 0