How would we prove this? I am familiar with deriving trig but not this. Can someone please shed some light and show me step by step on how to do this? Thank you.
You should know that:
$\displaystyle cos(A+B) = cosA.cosB - sinA.sinB$
$\displaystyle cos(A-B)= cosA.cosB + sinA.sinB$
Subtract these two equations.
$\displaystyle cos(A+B) - cos(A-B) = ????$....(1)
Now let $\displaystyle u = A+B$ and $\displaystyle v= A-B$
substitute this into equation (1)
Thank you for your help.
After I plugged that into the equation again the (cos a+b * cos a-b) cancels out and we are left with (-sin a + b sin a - b) - (sin a + b sin a - b) which should equal to -2 sin a + b sin a - b.
Now what would we do next? And what about the 2 in the denominator of the right hand side of the equation?
okay.
So you have:
$\displaystyle cos(A+B) - cos(A-B) = -2 sinA.sinB $.....(1)
As I said above:
let $\displaystyle u = A+B$ and let $\displaystyle v = A-B$......(2)
Clearly, the left side of the equation them becomes: $\displaystyle cosu - cosv$
your right side has $\displaystyle -2 sinA.sinB$
Convert $\displaystyle u$ and $\displaystyle v$ from (2) in terms of A and B