(r)爆頭都挴唔到有冇人幫我 amath黎的

y×sinA＝x×cosA
Let∠SPQ be A ,∠SQR be A ,QR be x ,PQ be y ,QS⊥PR ,ie ∠PSQ＝∠RSQ
∴QS=PQ sin ∠QPS=QR cos ∠RQS
∴∠PQS＝180°－∠PSQ－∠SPQ(∠ sum of ⊿)＝90°-A
∠PQR=∠PQS＋∠SQR=90°-A+A=90°
∴PR^2=PQ^2+QR^2=x^2+y^2
∴sin∠RPQ=QR/PR
∴sinA=x÷√(x^2+y^2)----3
∴cos∠RPQ=PQ÷PR
∴cosA=y÷√(x^2+y^2)----3

2007-04-28 22:16:07 補充：
sinA=x÷√(x^2 y^2) x=sinA × √(x^2 y^2)------5cosA=y÷√(x^2 y^2) y=cosA× √(x^2 Y^2)-----6sub 5 and 6 into 2 (sin^2 A)/[cosA × √(x^2 y^2)]^2 (cos^2 A)/[sinA × √(x^2 y^20]^2=2/(x^2 y^2)

2007-04-28 22:16:20 補充：
(sinA÷cosA)^2 (cosA÷sinA)^2＝2(sin^2 A)^2 (cos^2 A)=2(sin^2 A)(cos^2 A)(sin^2 A－cos^2 A)^2=0sin^2 A－(1－sin^2 A)＝02sin^2 A＝1sin^2 A=1÷2sinA＝√2÷2A＝45°

2007-04-28 22:18:20 補充：
sorr 個 號出不到