\[ \sum_{ k=0 }^{ \infty }\sum_{ i=0 }^{ \infty } \dfrac{ { (-1) }^{ i+k } }{ \left( 2k+1 \right)!(2i)!(i+k+1) } = \sin^b(a) \]
If the above equation is true for natural numbers \( a \) and \( b \). The find the value of \(a+b\).

**Note:** Angles are measured in radian.