Question

Hi, everyone.
I have this mathematical problem to prove:
sin(a)+sin(b)+sin[(a+b)/2] = 4cos(a/2)cos(b/2)sin[(a+b)/2]

Answer 1

Looking at the LEFT side of the thing, you can add two sines by using Simpson's formula:

sin(a)+sin(b)=2sin[(a+b)/2]

IMO that looks promising ;)

Now I'll look at the right side...

Answer 2

thanks :)

Answer 3

Sorry, that Simpson formaula didn't came out right:

sin(a)+sin(b)=2sin[(a+b)/2]cos[(a-b)/2]

Answer 4

oh, ok i'll re-write that :)

Answer 5

So now I've got on the left hand side:

2sin[(a+b)/2]cos[(a-b)/2]+sin[(a+b)/2]

Now sin[(a+b)/2] can be factored out:

sin[(a+b)/2](2cos[(a-b)/2]+1)

Answer 6

where did your 2come from 2sin[(a+b)/2]
and when you move it to the right it would be minus

Answer 7

The sin[(a+b)/2] is also a factor on the right hand side, so now we have to prove:

(2cos[(a-b)/2]+1=4cos(a/2)cos(b/2)

Do you see what I mean?

Answer 8

I see it :)

Answer 9

OK, let's focus on 2cos[(a-b)/2]+1

You probably know the formula cos(a-b)=cos(a)cos(b)+sin(a)sin(b), so:

\[2\cos \frac{ a-b }{ 2 }+1=2\cos \left( \frac{ a }{ 2 }-\frac{ b }{ 2 } \right)+1=2\cos \frac{ a }{ 2 }\cos \frac{ b }{ 2 }+2\sin \frac{ a }{ 2}\sin \frac{ b }{ 2 }+1\]

Answer 10

Ok, i get it. 'till there, what should i do next?

Answer 11

Look at the right hand side. Only \[4\cos \frac{ a }{ 2 }\cos \frac{ b }{ 2 }\]left there, remember?
Now it looks a lot like what we have found on the left hand side, in fact it is\[2\cos \frac{ a }{ 2 }\cos \frac{ b }{ 2 }+2\cos \frac{ a }{ 2 }\cos \frac{ b }{ 2 }\]So if I compare these two sides, I think I have to prove that\[2\sin \frac{ a }{ 2 }\sin \frac{ b }{ 2 }+1=2\cos \frac{ a }{ 2 }\cos \frac{ b }{ 2 }\]

Answer 12

Looking at this, I'm thinking that this won't hold for every a and b, I'm afraid...

Answer 13

ok, but it's fine ;) thx a lot!

Answer 14

YW!