# Double Angle Question

• Mar 7th 2009, 01:14 AM
noobonastick
Double Angle Question
Show that sin S - sin T = 2.cos (S + T / 2) .sin (S - T / 2)

(By the way, am I right in saying that you are not allowed to expand the RHS when the question asks to 'show'?)

Thanks !
• Mar 7th 2009, 01:33 AM
running-gag
Hi

You are allowed to expand the RHS using the usual trig identities cos(a+b) and sin(a-b)
• Mar 7th 2009, 11:56 AM
noobonastick
I still don't know how to do it. (Crying)
• Mar 7th 2009, 12:09 PM
running-gag
On the right side

$\cos \left(\frac{S}{2}+\frac{T}{2}\right) = \cos \frac{S}{2} \cos \frac{T}{2} - \sin \frac{S}{2} \sin \frac{T}{2}$

$\sin \left(\frac{S}{2}-\frac{T}{2}\right) = \sin \frac{S}{2} \cos \frac{T}{2} - \cos \frac{S}{2} \sin \frac{T}{2}$

Multiply both to get the RHS

On the left side

$\sin S - \sin T = 2 \sin \frac{S}{2} \cos \frac{S}{2} - 2 \sin \frac{T}{2} \cos \frac{T}{2}$