The expression \(\cos(x) + \cos(3x) + \cos(7x) + \cos(9x)\) can be written in the equivalent form \(a \cos(bx)\cos(cx)\cos(dx)\)
for some positive integers \(a\), \(b\), \(c\), and \(d\).
Find \(a + b + c + d\).
Formula:
\(\boxed{\cos(x)+\cos(y)=2\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right) }\)
\(\begin{array}{|rcll|} \hline && \mathbf{\cos(x) + \cos(3x) + \cos(7x) + \cos(9x)} \\\\ &=& \Big(\cos(9x) + \cos(x)\Big) + \Big(\cos(7x) + \cos(3x)\Big) \\\\ &=& 2\cos\left(\dfrac{9x+x}{2}\right)\cos\left(\dfrac{9x-x}{2}\right) + 2\cos\left(\dfrac{7x+3x}{2}\right)\cos\left(\dfrac{7x-3x}{2}\right) \\\\ &=& 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{8}{2}x\right) + 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{4}{2}x\right) \\\\ &=& 2\cos(5x)\cos(4x) + 2\cos(5x)\cos(2x) \\\\ &=& 2\cos(5x)\Big(\cos(4x) + \cos(2x)\Big) \\\\ &=& 2\cos(5x)*2\cos\left(\dfrac{4x+2x}{2}\right)\cos\left(\dfrac{4x-2x}{2}\right) \\\\ &=& 4\cos(5x)\cos\left(\dfrac{6}{2}x\right)\cos\left(\dfrac{2}{2}x\right) \\\\ &=&\mathbf{ 4\cos(5x)\cos(3x)\cos(x) } \qquad a=4,\ b=5,\ c=3,\ d=1 \\ \hline && \mathbf{a+b+c+d } \\ &=& 4+5+3+1 \\ &=& \mathbf{13} \\ \hline \end{array}\)