Sign equality in proof of (sin A +sin B) trig identitiy

[nomadreid]

Homework Statement

To show that sin A + sin B = +2|sin ((A+B)/2)*cos((A-B)/2)| is easy. However, it is not clear how to remove the absolute value signs to give the valid identity sinA + sin B = 2(sin ((A+B)/2)*cos((A-B)/2)) without having to go through many cases.

Homework Equations

To prove the version with the absolute value is a straightforward application on the RHS of the half-angle identities and the angle addition/subtraction identities, whereupon one gets
+√(sinA+sinB)².

The Attempt at a Solution

I started with making a table of the different cases, some of which are straightforward. But the number of cases that must be examined gets to quickly be rather large, so I figure that there is an easier way to go about it. All help would be appreciated.

 

[Dick]

Homework Statement

To show that sin A + sin B = +2|sin ((A+B)/2)*cos((A-B)/2)| is easy. However, it is not clear how to remove the absolute value signs to give the valid identity sinA + sin B = 2(sin ((A+B)/2)*cos((A-B)/2)) without having to go through many cases.

Homework Equations

To prove the version with the absolute value is a straightforward application on the RHS of the half-angle identities and the angle addition/subtraction identities, whereupon one gets
+√(sinA+sinB)².

The Attempt at a Solution

I started with making a table of the different cases, some of which are straightforward. But the number of cases that must be examined gets to quickly be rather large, so I figure that there is an easier way to go about it. All help would be appreciated.

I would prove it by adding sin(x+y) and sin(x-y) and then expanding them by addition formulas. Then put x=(A+B)/2 and y=(A-B)/2. Then the sign ambiguity never comes up.

 

[nomadreid]

Thanks a million, Dick. Not only does the sign ambiguity never come up that way, but this is easier. Very elegant.