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Deriving sinA+sinB formula
By Anonymous on Tuesday, January 9, 2001 - 09:41 am:

Hi,

Could someone tell me how this trig identity was derived?

Sin A + Sin B = 2Sin((A+B)/2).Cos((A-B)/2)

Was it using something using like C = A + B?
Thank you.

By Anonymous on Tuesday, January 9, 2001 - 11:52 am:

I can't tell you how it was derived, but I can prove it if that's any help:

Double angle formulas:
cos(X+Y)=cosX.cosY-sinX.sinY
sin(X+Y)=cosX.sinY+cosY.sinX

These of course imply that:
cos((X-Y)/2)=cos(X/2).cos(Y/2)+sin(X/2).sin(Y/2)
sin((X+Y)/2)=cos(X/2).sin(Y/2)+cos(Y/2).sin(X/2)

If we look at the R.H.S of your identity:
2Sin((A+B)/2).Cos((A-B)/2)

From the above this is equal to:
2[(cos(A/2).sin(B/2)+cos(B/2).sin(A/2)).(cos(A/2).cos(B/2)+sin(A/2).sin(B/2))]

Multiplying out and factorising we get:
2[sin(A/2).cos(A/2).(cos2(B/2)+sin2(B/2)) + sin(B/2).cos(B/2).(cos2(B/2)+sin2(B/2))]

Using the identity:
(cos2X+sin2X = 1

The right hand side becomes:
2sin(A/2).cos(A/2) + 2sin(B/2).cos(B/2)

From the double angle formula for sin(X) = sin(X/2+X/2), we see that the RHS = sin A + sin B

Thus the RHS = LHS and the proof is complete.

By Anonymous on Tuesday, January 9, 2001 - 01:12 pm:

Thank you.

however I thought that you could do it in a more simple (but not saying its better than your method) way of doing this by subbing something along the lines of I remember C = A + B / 2 and D = A - B / 2. If you see what I mean. I hope I am making sense.

By Brad Rodgers (P1930) on Wednesday, January 10, 2001 - 01:02 am:

Are you familiar with

sin(M+N)+sin(M-N)=2sin(M)cos(N)

Let

A=M+N
B=M-N

subbing in for the N in one of the equations and solving for M,

M=(A+B)/2

Then,

N=(A-B)/2

Using this, you should be able to find the derivation you're looking for.

If you're not familiar with the starting formula, it is relatively easy to prove, so just post back.

Brad

By Anonymous on Wednesday, January 10, 2001 - 11:40 am:

Thanks Brad for your help!