# Stuck :s

1. Oct 16, 2005

### mohlam12

Hey
So, I have to show that in a triangle, we have : (A, B, C are the angles of that triangle)

sinB+sinC-sinA=4cos(A/2)sin(B/2)sin(C/2)

okay, here is what I got to, then I got stuck
i got it equal to:

sinB+sinC-sinA = 2cos(A/2)cos(B/2)cos(C/2)+2cos(A/2)sin(B/2)sin(C/2)-2sin(B/2)sin(C/2)

I don't know what to do next! Any hints?

Last edited: Oct 16, 2005
2. Oct 16, 2005

### Fermat

You know that A + B + C = π

or,

A/2 = π/2 - (B+C)/2

hence,

sin(A/2) = sin(π/2 - (B+C)/2) = cos((B+C)/2)

and,

cos(A/2) = cos(π/2 - (B+C)/2) = sin((B+C)/2)

I wouldn't use these identities in the expression you already have, but rather start from the beginning again, keeping these identies in mind.

3. Oct 16, 2005

### mohlam12

yes, that's what I did to get to the equation above. But I dont know what to do after...! So it can be equal to 4cos(A/2)sin(B/2)sin(C/2)

Last edited: Oct 16, 2005
4. Oct 16, 2005

### Fermat

sinB + sinc - sinA =
2sin((B+C)/2)cos((B-C)/2) - 2sin(A/2)cos(A/2)

now use those identities

5. Oct 16, 2005

### mohlam12

yup, i used that before, and didn't get to anything good.
Here is what I could do with it:

sinB+sinC-sinA =
2cos(A/2)cos(B/2)cos(C/2)+2cos(A/2)sin(B/2)sin(C/2)+2sin(A/2)cos(B/2)cos(C/2)-2sin(A/2)sin(B/2)sin(C/2)

6. Oct 16, 2005

### Fermat

sinB + sinc - sinA =
2sin((B+C)/2)cos((B-C)/2) - 2sin(A/2)cos(A/2)

using the identities, we get,

2cos(A/2).cos((B-C)/2) - 2sin(A/2).cos(A/2)
2cos(A/2){cos((B-C)/2) - sin(A/2)}

using the identities again,

2cos(A/2){cos((B-c)/2) - cos((B+C)/2)}

now use the formula for the difference of two cosines,

2cos(A/2){2sin(B/2)sin(C/2)}
4cos(A/2)sin(B/2)sin(C/2)
===================

7. Oct 16, 2005

### mohlam12

OH YES! Sorry I didnt notice that I should use the formula for the difference of two cosines... Thanks a lot!