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Prove the following identities
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\[\cos2 \theta + \cos 4 \theta + \cos 6 \theta + \cos 12 \theta = 4 \cos 3 \theta \cos 4 \theta \cos 5 \theta\]
We would have to use this formula repeatedly: \[\large{\cos x + \cos y = 2\cos(\cfrac{x+y}{2}) \cos(\cfrac{x-y}{2})}\]
ok..
Use this formula for: \[\large{\cos{2\theta} + \cos{6\theta},~~\cos{4\theta} + \cos{12\theta}}\]
I am solving for the first sum and you solve for the second one
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I am pretty sure there is a shortcut formula for this type of sum though :)
But lets move on...
\[\large{\cos{6\theta} + \cos{2\theta}}\] \[\large{=2\cos(\cfrac{6\theta+2\theta}{2}) \cos(\cfrac{6\theta - 2\theta}{2})}\]
\[\large{= 2\cos{4\theta}\cos{2\theta}}\]
|dw:1405585164945:dw|
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