1. ## Proving Trigonometry function

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.

2. You can expand the RHS using compound angle formula.

3. Originally Posted by florx

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)

4. 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?

5. Originally Posted by florx

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