Question

Simplify the following.

Answer 1

$\bf cos({{ \alpha}})-cos({{ \beta}})=-2sin\left(\cfrac{{{ \alpha}}+{{ \beta}}}{2}\right)sin\left(\cfrac{{{ \alpha}}-{{ \beta}}}{2}\right) \\\\\\ sin({{ \alpha}})+sin({{ \beta}})=2sin\left(\cfrac{{{ \alpha}}+{{ \beta}}}{2}\right)cos\left(\cfrac{{{ \alpha}}-{{ \beta}}}{2}\right) \\\\\\ \textit{also recall the symmetry identity of }sin(-\theta )=-sin(\theta )\\\\ -------------------------------$

$\bf \cfrac{cos(3x)-cos(7x)}{sin(7x)+sin(3x)}\implies \cfrac{-2sin\left( \frac{3x+7x}{2} \right)sin\left( \frac{3x-7x}{2} \right)}{2sin\left( \frac{7x+3x}{2} \right)cos\left( \frac{7x-3x}{2} \right)} \\\\\\ \cfrac{-2sin\left( \frac{10x}{2} \right)sin\left( \frac{-4x}{2} \right)}{2sin\left( \frac{10x}{2} \right)cos\left( \frac{4x}{2} \right)}\implies \cfrac{-\underline{2sin\left( 5x \right)} sin\left( -2x \right)}{\underline{2sin\left( 5x \right)} cos\left( 2x \right)}$

$\bf \cfrac{-sin(-2x)}{cos(2x)}\implies \cfrac{-[-sin(2x)]}{cos(2x)}\implies \cfrac{sin(2x)}{cos(2x)}\implies tan(2x)$