Solution assignment 02 Addition and subtraction formulas

Show:

$\sin(x)+\sin(y)=2\sin[\displaystyle\frac{1}{2}(x+y)]\cos[\displaystyle\frac{1}{2}(x-y)]$

We know:

$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

$\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)$

Adding both sides yields:

$\sin(x+y)+\sin(x-y)=2\sin(x)\cos(y)$

Assume:

$u=x+y$

$v=x-y$

Then we express $x$ and $y$ in $u$ and $v$ , and we get:

$x=\displaystyle\frac{1}{2}(u+v)$

$y=\displaystyle\frac{1}{2}(u-v)$

Then the formula becomes:

$\sin(u)+\sin(v)=2\sin[\displaystyle\frac{1}{2}(u+v)]\cos[\displaystyle\frac{1}{2}(u-v)]$

When we replace $u$ by $x$ and $v$ by $y$ (what's in a name?), then we get the formula.