F.4 trigonometry---3

用戶36732

2010-03-11 3:49 am

http://lsforum.net/upload/users/public/k53755MATH----i114.jpg

In the figure, ∠AOB=π/3, P is a variable point such that OP=10 and ∠AOP=θ(0≦θ≦π/3). M and N are the feet of perpendiculars from P to OA and OB respectively.
(a)
(1) Find OM and ON in terms of θ.
(2) Hence, find the maximum value of OM+ON.
(b)
(1) Find PM and PN in terms of θ.(2) Hence, find the maximum value of PM+PN.

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回答 (1)

用戶2053

2010-03-11 4:55 am

✔ 最佳答案

a1) OM/OP = cos θ
OM = OP cos θ = 10 cos θ
ON / OP = cos (π/3 - θ)
ON = 10 cos (π/3 - θ)
a2) OM+ON
= 10 cos θ + 10 cos (π/3 - θ)
= 10 (cos θ + cos (π/3 - θ))
= 10 [2cos (θ + (π/3 - θ))/2 cos (θ - (π/3 - θ))/2]
= 20 cos (π/6) cos(θ - π/6)
= (20 √3/2) cos(θ - π/6)
= (10√3) cos(θ - π/6) ,
Since Max. cos x = 1 ,
When θ = π/6 , Max. OM+ON = (10√3) cos 0 = 10√3

b1)PM / OP = sinθ , PM = 10 sinθ
PN / OP = sin(π/3 - θ) , PN = 10 sin(π/3 - θ)
PM + PN = 10 (sinθ + sin(π/3 - θ))
= 10(2sin (θ + π/3 - θ)/2 cos ((θ - (π/3 - θ))/2)
= 20 (sin π/6)(cos (θ - π/6))
= 20 (1/2) (cos (θ - π/6))
= 10 cos(θ - π/6)
Max. PM + PN = 10 , (when cos(θ - π/6) = 1 , i.e. θ = π/6)

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