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a-maths~~~~~~~
發問:
在?ABC,設a/sinA=b/sinB=c/sinC=k,其中A,B和C是?ABC的內角。 (a)証明a/sinA=(a+b+c)/(sinA+sinB+sinC) (b)(i)利用性等式sin2θ≡2sinθcosθ,証明sinC=2sin[(A+B)/2]cos[(A+B)/2]。 (ii)由此,証明sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)。 (c)已知a+b+c=2m。利用(a)和(b)小題結果,証明a=msin(A/2)sec(B/2)sec(C/2)。
最佳解答:
a)R.H.S. = (a+b+c)/(sinA+sinB+sinC) = (a+b+c)/(a/k+b/k+c/k) = (a+b+c)/(1/k)(a+b+c) = 1/(1/k) = k = a/sinA = L.H.S. bi)L.H.S.=sinC =sin2(C/2) =2sin(C/2)cos(C/2) =2sin[[180*-(A+B)]/2]cos[[180*-(A+B)]/2] =2sin[90*-(A+B)/2]cos[90*-(A+B)/2] =2sin[(A+B)/2]cos[(A+B)/2] =R.H.S. ii)L.H.S.= sinA+sinB+sinC = 2sin[(A+B)/2]cos[(A-B)/2]+2sin[(A+B)/2]cos[(A+B)/2] = 2sin[(A+B)/2][cos[(A-B)/2]+cos[(A+B)/2]] = 2sin[(A+B)/2][2cos(A/2)cos(B/2)] = 4sin[(180*-C)/2][cos(A/2)cos(B/2)] = 4sin(90*-C/2)cos(A/2)cos(B/2) = 4cos(A/2)cos(B/2)cos(C/2) = R.H.S. c) a/sinA=(a+b+c)/(sinA+sinB+sinC) a/sinA=2m/[4cos(A/2)cos(B/2)cos(C/2)] a = 2msinA/[4cos(A/2)cos(B/2)cos(C/2)] = 4msin(A/2)cos(A/2)/[4cos(A/2)cos(B/2)cos(C/2)] = msin(A/2)sec(B/2)sec(C/2)
其他解答:
(a) a/sinA = k 根據比例的性質 (a+b+c)/(sinA+sinB+sinC) = k so a/sinA = (a+b+c)/(sinA+sinB+sinC) (b)(i) sinC = 2sin(C/2)cos(C/2) C = 180 - A - B =2sin[(A+B)/2]cos[(A+B)/2] (ii) sinA + sinB + sinC = 2sin[(A-B)/2]cos[(A-B)/2] + 2sin[(A+B)/2]cos[(A+B)/2] = sin[(A+B)/2]{cos[(A-B)/2] + cos[(A+B)/2]} =4cos(A/2)cos(B/2)cos(C/2) (c) a/sinA = (a+b+c)/(sinA+sinB+sinC) a=sinA(a+b+c)/(sinA+sinB+sinC) = 4msin(A/2)cos(A/2) / 4cos(A/2)cos(B/2)cos(C/2) = msin(A/2)sec(B/2)sec(C/2)
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