標題:
a-maths!!!!!!
http://img.photoamp.com/pa/07/08/14/1H06UIPT.jpg 點做a?
最佳解答:
a)SR = a sinθ OQ = a sinθ/ tan 2α OR = a cosθ QR = PS = a cosθ- a sinθ/ tan 2α = ( a sin 2αcosθ- a sinθcos 2α) / sin 2α = a sin ( 2α- θ) / sin 2α b)K = SR x PS = a sinθ{ a sin ( 2α- θ) / sin 2α} = a2 sinθsin ( 2α- θ) / sin 2α = (a2 / 2 sin2α){ 2 sinθsin ( 2α- θ) } = ( a2 / 2 sin 2α){ cos ( 2θ- 2α) – cos 2α} = ( a2 / 2 sin 2α){ cos ( 2α- 2θ) – cos 2α} K的極值(當cos ( 2α- 2θ) = 1): K = ( a2 / 2 sin 2α)( 1 – cos 2α) = ( a2 / 4 sinαcosα)( 2 sin2α) = a2 tanα/ 2
其他解答:
(a) In triangle SOR, sin α = SR/a SR = a sin α ∠OPQ = 180 - 90 -2α = 90-2α ∠OPS = 90+90-2α = 180-2α In triangle OPS, a/sin (180-2α) = PS/sin(2α-θ) a/sin2α=PS/sin(2α-θ) Hence PS = asin(2α-θ)/sin2α
a-maths!!!!!!
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發問:http://img.photoamp.com/pa/07/08/14/1H06UIPT.jpg 點做a?
最佳解答:
a)SR = a sinθ OQ = a sinθ/ tan 2α OR = a cosθ QR = PS = a cosθ- a sinθ/ tan 2α = ( a sin 2αcosθ- a sinθcos 2α) / sin 2α = a sin ( 2α- θ) / sin 2α b)K = SR x PS = a sinθ{ a sin ( 2α- θ) / sin 2α} = a2 sinθsin ( 2α- θ) / sin 2α = (a2 / 2 sin2α){ 2 sinθsin ( 2α- θ) } = ( a2 / 2 sin 2α){ cos ( 2θ- 2α) – cos 2α} = ( a2 / 2 sin 2α){ cos ( 2α- 2θ) – cos 2α} K的極值(當cos ( 2α- 2θ) = 1): K = ( a2 / 2 sin 2α)( 1 – cos 2α) = ( a2 / 4 sinαcosα)( 2 sin2α) = a2 tanα/ 2
其他解答:
(a) In triangle SOR, sin α = SR/a SR = a sin α ∠OPQ = 180 - 90 -2α = 90-2α ∠OPS = 90+90-2α = 180-2α In triangle OPS, a/sin (180-2α) = PS/sin(2α-θ) a/sin2α=PS/sin(2α-θ) Hence PS = asin(2α-θ)/sin2α
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