Who is asking: Student
Level: Secondary
Question:
Prove:
sin A + sin B = 2sin(A+B/2)cos(A-B/2)
cos A - cos B = -2sin(A+B/2)sin(A-B/2)
cos A + cos B = 2cos(A+B/2)cos(A-B/2)
sin A - sin B = 2cos(A+B/2)sin(A+B/2)
Please help im almost there with these babies and it's very frustrating!
Hi,
I will illustrate how to derive the first expression. To do it I am going to assume that you know that
sin(P + Q) = sin(P)cos(Q) + cos(P)sin(Q) (equation 1)
and that
sin(-P) = -sin(P) and
cos(-P) = cos(P)
From these I can derive
sin(P - Q) = sin(P + (- Q)) = sin(P)cos(-Q) + cos(P)sin(-Q) so
sin(P - Q) = sin(P)cos(Q) - cos(P)sin(Q) (equation 2)
Adding equation 1 and equation 2 I get
sin(P + Q) + sin(P - Q)
= sin(P)cos(Q) + cos(P)sin(Q) + sin(P)cos(Q) - cos(P)sin(Q)
thus
sin(P + Q) + sin(P - Q) = 2 sin(P)cos(Q)
If you now let P = A + B/2 and Q = A - B/2 you will se that this is exactly the expression in your first problem.
The other can be derived in a similar fashion.
Penny