# Thread: Deriving a compound angle formula..

1. ## Deriving a compound angle formula..

The compound angle formulas can be used to derive many other formulas. One of these is the formula

a) Show that the formula is valid for x = 2pi/3 and y = pi/3. (I did this, both sides were equal to square root of 3)

b) Use appropriate equivalent trigonometric expressions to derive a similar formula for sin x - sin y.

How would I show my work as to get to that answer for b? I don't really know what b is even asking.

2. You will use the fact that $\sin y$ is an odd function, i.e., $- \sin y = \sin (-y)$.

3. Simple question... are these two the same thing?:

sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)

and

sin x - sin y = 2 sin ((x-y) / 2) cos ((x+y) / 2)

if they are I think I got it

4. You are asking is $u \cdot v = v \cdot u$. You understand that it doesn't matter in which order I multiply numbers yes?

5. Just making sure

b) Use appropriate equivalent trigonometric expressions to derive a similar formula for sin x - sin y.

So is all I have to do for that is come up with sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)? Do I have to do any calculations to prove that it works?

6. No, you just need to use trig identities to show it is equal.

7. Hello, kmjt!

If they want us to derive "from scratch", it requires some acrobatics.

b) Use appropriate equivalent trigonometric expressions to derive
a similar formula for: . $\sin x - \sin y$

Answer: . $\sin x - \sin y \:=\:2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$
Note that: . $\begin{array}{ccc}x &=& \dfrac{x+y}{2} + \dfrac{x-y}{2} \\ \\[-3mm] y &=& \dfrac{x+y}{2} - \dfrac{x-y}{2} \end{array}$

Then:

. . $\sin x \;=\; \sin\left(\frac{x+y}{2} + \frac{x-y}{2}\right)$ . $=\; \sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\;\;[1]$

. . $\sin y \;=\; \sin\left(\frac{x+y}{2} - \frac{x-y}{2}\right)$ . $=\; \sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) - \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\;\;[2]$

Subtract [2] from [1]: . $\sin x - \sin y \;=\;2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$

8. You guys have both been very helpful. Thanks!

9. I accept with information:sin x - sin y = 2 cos ((x+y) / 2) sin ((x-y) / 2)
and sin x - sin y = 2 sin ((x-y) / 2) cos ((x+y) / 2)
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