The answer:
the full question is as follow:
if A+B-C=3pi, then find sinA+sinB-sinC
first, the main formula of sine and cosine are:
sinC = 2sin(C/2)cos(C/2)
sinA +sinB = 2sin[(A+B)/2]cos[(A-B)/2]
therefore:
sinA+sinB-sinC = 2sin[(A+B)/2]cos[(A-B)/2] - 2sin(C/2)cos(C/2)
sin[(A+B)/2] = cos(C/2)
2sin(C/2)cos(C/2) = cos[(A+B)/2
and
A+B-C=3 pi implies A+B =3 pi + C, so
cos[(A+B)/2] = cos [3 pi/2 + C/2]
and with the equivalence cos (3Pi/2 + X) = sinX
sinA+sinB-sinC = cos(C/2)+ sin(C/2)
42%7=6. 6 times 5 is 30. 30 cakes in 5 days.
Answer:
thats right
Step-by-step explanation:
Answer:
to what??????
Step-by-step explanation: ???????
