Question

If A+B+C = then prove that: n sin A cos A + sin B cos B + sin C cos C=2 sin A sin B sin C​

Answer 1

Let a, b, c be the sides that oppose angles A, B, C, respectively. By the Law of Sines,

asinA=bsinB=csinC=d

where d is the circumdiameter of the triangle. If we conveniently scale the triangle so that d=1, then we can say simply that

a=sinAb=sinBc=sinC

This is a common simplification technique, as it nicely blurs the distinction between edges and angles, giving us things like this wonderfully-symmetric area formula:

Area=12abc(=12absinC=12acsinB=12bcsinA)

(When you have this mindset, you can't look at the expression "sinA+sinB+sinC" and not think, "That's perimeter!" ... and then you find yourself pursuing proof approaches like this one.) In what follows, I'll continue to use "a", "b", "c", because they're more compact than "sinA", etc, but you should read them as "sinA", etc.