1條A. Maths Compound Angles問題

(a)
OQ + OR = x cos(π/4 - θ) + x cos θ
d(OQ + OR)/dθ = x sin(π/4 - θ) - x sin θ
d(OQ + OR)/dθ = 0
　when sin(π/4 - θ) = sin θ
　　　　　 π/4 - θ = θ 　　, as 0 ≦ θ ≦ π/4
　　　　　　　　θ = π/8
(cos π/8)^2
　= (1/2) (1 + cos π/4)
　= (1/2) ( 1 + (√2)/2 )
　= (1/4) (2 + √2)
cos π/8 = (1/2) √(2 +√2)

Thus,
maximum value of OQ + OR
　= x cos(π/4 - π/8) + x cos π/8
　= 2x cos π/8
　= x √(2 +√2).

(b)
Area of △OQR
　= (1/2) OQ • OR sin π/4
　= [1 / (2√2)] [x cos(π/4 - θ)] • [x cos θ]
　= [1 / (2√2)] x^2 • (1/2) [ cos ( θ + (π/4 - θ) ) + cos ( θ - (π/4 - θ) ) ]
　= [1 / (4√2)] x^2 • [ cos π/4 + cos ( 2θ - π/4 ) ]
d(Area of △OQR)/dθ
　= [1 / (4√2)] x^2 • [ 0 - 2 sin ( 2θ - π/4 ) ]
d(Area of △OQR)/dθ = 0
　when sin ( 2θ - π/4 ) = 0
　　　　　　2θ - π/4 = 0 　　, as 0 ≦ θ ≦ π/4
　　　　　　　　 θ = π/8
Thus,
maximum area of △OQR
　= [1 / (4√2)] x^2 • [ cos π/4 + cos ( 2 (π/8) - π/4 ) ]
　= [1 / (4√2)] x^2 • ( 1/√2　+ 0 )
　= (x^2) / 8.

2007-01-19 20:59:52 補充：
In (b), it is "quite intuitive" that θ = π/8 gives max. area of △OQR, because for max. area, △OQR must be an isoceles triangle. But of course a proof with d(Area of △OQR)/dθ = 0 is more rigorous and is the expected answer in CE Add. Maths. exam.

(b) First of all, length of OQ = x cos (θ-π/4) and length of OR = x cos θ
Hence area of △OQR is given by: (1/2) × OQ × OR × sin π/4
= (1/2)[x cos (θ-π/4)](x cos θ) × sin π/4
= (1/2)x2 [cos (θ-π/4) cos θ] × sin π/4
= (1/2)x2 (1/2) [cos (2θ-π/4)/2 + cos (θ-θ-π/4)/2] × sin π/4 by using the formula cos Acos B = (1/2){cos [(A+B)/2] + cos[(A-B)/2]}
= (1/4)x2 [cos (θ - π/8) + cos (-π/8)] × sin π/4
= (1/4)x2 [cos (π/8) + cos (θ - π/8)] × sin π/4
∴ We can see that the area of △OQR is maximum when θ = π/8 since cos 0 = 1.
Hence max. area = (1/4)x2 [cos (π/8) + cos (- π/8)] × sin π/4
= (1/2)x2 cos (π/8)sin π/4
= (1/2)x2 (1/√2) √{[cos (π/4) + 1]/2} by using the formula cos 2θ = 2cos2 θ - 1.
= (1/2√2)x2 √(1/2√2 + 1/2)
= (1/4)x2 √(1/2 + 1/√2)
= (1/4)x2 √[(2 + √2)/2]