using cosine rule prove sine rule
Answers
I think ultimately the Law of Cosines is the most fundamental one. So this is a pretty good question. Perhaps a more pointed question is: assuming the Law of Cosines, what else do we need to prove the Law of Sines? The answer seems to be just the Trig Pythagorean Theorem, cos2θ+sin2θ=1. That’s about the minimum possible that turns a cosine into a sine; the rest is algebra.
We’re given triangle ABC with sides a,b,c labeled in the usual way and we know the Law of Cosines is true:
c2=a2+b2−2abcosC
We can write this for each side:
a2=b2+c2−2bccosA
b2=a2+c2−2accosB
Let’s rearrange these and square. We’ll just work with the first one.
(c2−a2−b2)2=4a2b2cos2C
a4+b4+c4−2a2c2−2b2c2+2a2b2=4a2b2(1−sin2C)
2(a2c2+b2c2+a2b2)−(a4+b4+c4)=4a2b2sin2C=4a2b2c2sin2Cc2
2(a2c2+b2c2+a2b2)−(a4+b4+c4)4a2b2c2=sin2Cc2
When we write this for the other two angles, the left hand side will be the same, because the expression is symmetric, staying the same when we interchange any pair of sides. So we get
sin2Aa2=sin2Bb2=sin2Cc2
That shows the squared form of the Law of Sines from the Law of Cosines. For triangle angles the sines are always positive, as are the lengths, so we can unambiguously take the square root of these equations and get
sinAa=sinBb=sinCc
Answer:
sine rule proved above.
