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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 $a + b + c + d.$

 Mar 31, 2020

Best Answer 

 #1
avatar+25597 
+2

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:

\(\boxed{\cos(x)+\cos(y)=2\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right) }\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(x) + \cos(3x) + \cos(7x) + \cos(9x)} \\\\ &=& \Big(\cos(9x) + \cos(x)\Big) + \Big(\cos(7x) + \cos(3x)\Big) \\\\ &=& 2\cos\left(\dfrac{9x+x}{2}\right)\cos\left(\dfrac{9x-x}{2}\right) + 2\cos\left(\dfrac{7x+3x}{2}\right)\cos\left(\dfrac{7x-3x}{2}\right) \\\\ &=& 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{8}{2}x\right) + 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{4}{2}x\right) \\\\ &=& 2\cos(5x)\cos(4x) + 2\cos(5x)\cos(2x) \\\\ &=& 2\cos(5x)\Big(\cos(4x) + \cos(2x)\Big) \\\\ &=& 2\cos(5x)*2\cos\left(\dfrac{4x+2x}{2}\right)\cos\left(\dfrac{4x-2x}{2}\right) \\\\ &=& 4\cos(5x)\cos\left(\dfrac{6}{2}x\right)\cos\left(\dfrac{2}{2}x\right) \\\\ &=&\mathbf{ 4\cos(5x)\cos(3x)\cos(x) } \qquad a=4,\ b=5,\ c=3,\ d=1 \\ \hline && \mathbf{a+b+c+d } \\ &=& 4+5+3+1 \\ &=& \mathbf{13} \\ \hline \end{array}\)

 

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 Mar 31, 2020
 #1
avatar+25597 
+2
Best Answer

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:

\(\boxed{\cos(x)+\cos(y)=2\cos\left(\dfrac{x+y}{2}\right)\cos\left(\dfrac{x-y}{2}\right) }\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(x) + \cos(3x) + \cos(7x) + \cos(9x)} \\\\ &=& \Big(\cos(9x) + \cos(x)\Big) + \Big(\cos(7x) + \cos(3x)\Big) \\\\ &=& 2\cos\left(\dfrac{9x+x}{2}\right)\cos\left(\dfrac{9x-x}{2}\right) + 2\cos\left(\dfrac{7x+3x}{2}\right)\cos\left(\dfrac{7x-3x}{2}\right) \\\\ &=& 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{8}{2}x\right) + 2\cos\left(\dfrac{10}{2}x\right)\cos\left(\dfrac{4}{2}x\right) \\\\ &=& 2\cos(5x)\cos(4x) + 2\cos(5x)\cos(2x) \\\\ &=& 2\cos(5x)\Big(\cos(4x) + \cos(2x)\Big) \\\\ &=& 2\cos(5x)*2\cos\left(\dfrac{4x+2x}{2}\right)\cos\left(\dfrac{4x-2x}{2}\right) \\\\ &=& 4\cos(5x)\cos\left(\dfrac{6}{2}x\right)\cos\left(\dfrac{2}{2}x\right) \\\\ &=&\mathbf{ 4\cos(5x)\cos(3x)\cos(x) } \qquad a=4,\ b=5,\ c=3,\ d=1 \\ \hline && \mathbf{a+b+c+d } \\ &=& 4+5+3+1 \\ &=& \mathbf{13} \\ \hline \end{array}\)

 

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heureka Mar 31, 2020
 #2
avatar+112413 
+1

Very  nice, heureka  !!!!

 

 

cool cool cool

CPhill  Mar 31, 2020
 #3
avatar+25597 
+1

Thank you, CPhill !

 

laugh

heureka  Mar 31, 2020

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