We are asked to evaluate two trigonometric expressions: (7) $sin(10^{\circ}) - sin(70^{\circ}) + sin(130^{\circ})$ (8) $\frac{sin(55^{\circ}) sin(35^{\circ})}{cos(80^{\circ}) + cos(40^{\circ})}$

1. Problem Description

We are asked to evaluate two trigonometric expressions:

(7) $sin(10^{\circ}) - sin(70^{\circ}) + sin(130^{\circ})$

(8) $\frac{sin(55^{\circ}) sin(35^{\circ})}{cos(80^{\circ}) + cos(40^{\circ})}$

2. Solution Steps

(7) We have $sin(10^{\circ}) - sin(70^{\circ}) + sin(130^{\circ})$.

Since $sin(130^{\circ}) = sin(180^{\circ} - 130^{\circ}) = sin(50^{\circ})$, the expression becomes $sin(10^{\circ}) - sin(70^{\circ}) + sin(50^{\circ})$.

We can rewrite this as $sin(10^{\circ}) + sin(50^{\circ}) - sin(70^{\circ})$.

Using the sum-to-product formula, $sin(A) + sin(B) = 2 sin(\frac{A+B}{2}) cos(\frac{A-B}{2})$,

we have $sin(10^{\circ}) + sin(50^{\circ}) = 2 sin(\frac{10^{\circ} + 50^{\circ}}{2}) cos(\frac{10^{\circ} - 50^{\circ}}{2}) = 2 sin(30^{\circ}) cos(-20^{\circ})$.

Since $sin(30^{\circ}) = \frac{1}{2}$ and $cos(-20^{\circ}) = cos(20^{\circ})$, we get $2 (\frac{1}{2}) cos(20^{\circ}) = cos(20^{\circ})$.

Therefore, the expression becomes $cos(20^{\circ}) - sin(70^{\circ})$.

Since $sin(70^{\circ}) = cos(90^{\circ} - 70^{\circ}) = cos(20^{\circ})$, the expression is $cos(20^{\circ}) - cos(20^{\circ}) = 0$.

(8) We have $\frac{sin(55^{\circ}) sin(35^{\circ})}{cos(80^{\circ}) + cos(40^{\circ})}$.

Using the product-to-sum formula, $sin(A) sin(B) = \frac{1}{2} [cos(A-B) - cos(A+B)]$, we have

$sin(55^{\circ}) sin(35^{\circ}) = \frac{1}{2} [cos(55^{\circ} - 35^{\circ}) - cos(55^{\circ} + 35^{\circ})] = \frac{1}{2} [cos(20^{\circ}) - cos(90^{\circ})] = \frac{1}{2} [cos(20^{\circ}) - 0] = \frac{1}{2} cos(20^{\circ})$.

Using the sum-to-product formula, $cos(A) + cos(B) = 2 cos(\frac{A+B}{2}) cos(\frac{A-B}{2})$, we have

$cos(80^{\circ}) + cos(40^{\circ}) = 2 cos(\frac{80^{\circ} + 40^{\circ}}{2}) cos(\frac{80^{\circ} - 40^{\circ}}{2}) = 2 cos(60^{\circ}) cos(20^{\circ})$.

Since $cos(60^{\circ}) = \frac{1}{2}$, we have $2 (\frac{1}{2}) cos(20^{\circ}) = cos(20^{\circ})$.

Therefore, the expression becomes $\frac{\frac{1}{2} cos(20^{\circ})}{cos(20^{\circ})} = \frac{1}{2}$.

3. Final Answer

(7) 0

(8) $\frac{1}{2}$