a-maths~~~~~~~

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)

(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)