simplify... !!!

[tatoo5ma]

hello
i have to simplify this :

sin(a)sin(b-c) + sin(b)sin(c-a) + sin(c)sin(a-b)

I did simplify it and I got it equal to zero. But my problem is, I did too many steps in order to do that. Does anyone have a shorter way to simplify that ??

thx!

[TD]

Well, how did you do it? Perhaps there is no shorter way or maybe we can just point out an alternative somewhere in your work.

[tatoo5ma]

ok here,
i simply expanded the sin(b-c) and sin(c-a) and sin(a-b)
and finally i got:
sin(a)*(sin(b)cos(c)-sin(c)cos(b))+sin(b)*(sin(c)cos(a)-sin(a)cos(c))+sin(c)*(sin(a)cos(b)-sin(b)cos(a))
and i still expand that to get:
sin(a)sin(b)cos(c)-sin(a)sin(c)cos(b)+sin(b)sin(c)cos(a)-sin(b)sin(a)cos(c)+sin(c)sin(a)cos(b)-sin(c)sin(b)cos(a)
yap, everythin with the same color cancels out, well in fact, i am just wondering if there is another solution shorter than that. that u can notice from the beginnin! :uhh:

[TD]

I don't think this is 'that long', only 3 lines actually :smile:
Perhaps the lines are a bit long, but that was to be expected. I would've approached this about the same way, it seems the natural thing to do: using the sin(x-y) = sin(x)cos(y)-sin(y)cos(x) identity.

[tatoo5ma]

hmmm i see, thanks :)
i also have to show that, in a triangle <-> A+B+C=pi ;
sinB+sinC-sinA=4cos(A/2)sin(B/2)sin(C/2)
if someone can just give a hint r something to start with, because this one is really confusing me! :confused:
thanks!

[TD]

You can use the triangle fact by substituting a with pi-b-c. Remember that sin(pi-x) = sin(x) and that cos(pi-x) = -cos(x). Then you could try using double-angle formulas on the sines at the LHS to go to half angles as well.

[tatoo5ma]

ok i ll try that!

[tatoo5ma]

okay i was tryin that durin the past 20min, and i didnt get to what i want, however i got to know that tan(A/2)=(1-cos(b))/sin(b) and even with this, i didnt get closer to what i want. help would be apprciated :)

[TD]

Where did you get stuck? Also, when you change cos(a/2) in the RHS to cos((pi-b-c)/2), you can simplify that to sin(b/2 + c/2), perhaps that helps?

[tatoo5ma]

okay, here is what i got to:

-2sin(B/2-C/2)cos(B/2+C/2)+2sin(B/2)cos(B/2)+2sin(C/2)cos(C/2)

okay, i might replace -2sin(B/2-C/2)cos(B/2+C/2) by -sinB-sinV but that s not gonna help me any better!

and how can this above, be equal to 4cos(A/2)sin(B/2)sin(C/2)

PS: as u might have noticed, it's like 1:10am over here, and i dun wanna leave this problem for tomorrow cuz tomorrow i ll have more...so please, if u can give me good hints for that, so i can do it quicky...thhhanks

[stmoe]

I don't know if this helps at all, but I would actually start using law of sines / cosines and like , 2 sin(a)cos(a) = sin(2a) identity...

Sadly, I don't have the time to work on it right now.
Also .. you do realize what you originally gave was just an expression, so you shouldn't get it equal to anything until you start incorporating the sides in and using the A+B+C = pi data to substitute in and stuff