# proof

• Nov 18th 2006, 03:11 PM
lexmex1546
proof
i need to prove sinx-siny= 2 sin( (x - y)/2 ) cos( (x + y)/2 )
• Nov 18th 2006, 04:22 PM
putnam120
here is a simple say.

$\displaystyle \sin(a)+\sin(b)=\sin(x+y)+\sin(x-y)$ for some x and y.

after applying the sine addition and subtraction formulas,

$\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm\sin(y)\cos(x)$, you will end up with

$\displaystyle 2\sin(x)\cos(y)$ now note that we want this in terms of a and b so we have

$\displaystyle x=\frac{a+b}{2}$ and $\displaystyle y=\frac{a-b}{2}$ so using this we have

$\displaystyle \sin(a)+\sin(b)=2\sin\left(\frac{a+b}{2}\right)\co s\left(\frac{a-b}{2}\right)$
• Nov 19th 2006, 05:52 AM
Soroban
Hello, lexmex1546!

This is a rehash of putnam120's solution . . . with baby steps.

Quote:

Prove: .$\displaystyle \sin x - \sin y \:= \:2\sin\left(\frac{x - y}{2}\right)\cos\left(\frac{x + y}{2}\right)$

We have these two formulas:

. . $\displaystyle (1)\;\sin(a + b) \:=\:\sin a\cos b + \sin b\cos a$
. . $\displaystyle (2)\;\sin(a-b) \:=\:\sin a\cos b - \sin b\cos a$

Subtract (2) from (1): .$\displaystyle \sin(a+b) - \sin(a-b) \:=\:2\sin b\cos a\;\;(3)$

Let: .$\displaystyle \begin{Bmatrix}x\:=\:a+b \\ y\:=\:a-b\end{Bmatrix}\;\;\Rightarrow\;\;\begin{Bmatrix}a \:=\:\frac{x+y}{2} \\ b \:= \:\frac{x-y}{2}\end{Bmatrix}$

Then (3) becomes: .$\displaystyle \boxed{\sin x - \sin y \:=\:2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)}$