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How the following conversion is justified ?

$$\cos \biggl( \theta + \frac{2\pi}{3} \biggr) + \cos \biggl( \theta + \frac{4\pi}{3} \biggr) = \cos \biggl( \frac{\pi}{2} - \theta \biggr) - \cos \biggl( \frac{\pi}{3} +\theta \biggr) $$

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    It's not correct. Try replacing $\theta$ with $\frac{\pi}{4}$.2010-11-14
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    Hm thanks, I found this step in my module,it raise some doubts in me too.2010-11-14
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    Second and fourth terms are equal, while first and third obviously aren't (as functions). They probably have a typo somewhere.2010-11-14
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    I'm assuming the first fraction on the right hand side should be $\frac{\pi}{3}$, no $\frac{\pi}{2}$ and that you are missing a minus sign.2010-11-14

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First, notice that $$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1)-\sin(x)(0) = -\cos(x).$$ So, rewriting, using the above, and using that $\cos(-x)=\cos x$ you have: \begin{align*} \cos\left(\theta+\frac{2\pi}{3}\right) + \cos\left(\theta+\frac{4\pi}{3}\right) &= \cos\left(\left(\theta-\frac{\pi}{3}\right)+\pi\right) + \cos\left(\left(\theta+\frac{\pi}{3}\right)+\pi\right)\\ &= -\cos\left(\theta-\frac{\pi}{3}\right) - \cos\left(\theta+\frac{\pi}{3}\right)\\ &= -\cos\left(\frac{\pi}{3}-\theta\right) - \cos\left(\frac{\pi}{3}+\theta\right). \end{align*} So: the equality as written cannot hold: you would need $\cos(\frac{\pi}{2}+\theta) = - \cos(\frac{\pi}{3}-\theta)$ for all $\theta$, and plugging in $\theta=0$ shows this cannot hold. I suspect your $\frac{\pi}{2}$ should have been a $\frac{\pi}{3}$, and that you are missing a minus sign before the first term. If this is not the case, then please edit the question to give the equality you are actually trying to prove.

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    I found this step mentioned in my module while they are trying to solve a problem,this step happened to be wrong.However,I solved the (mother) problem using a slight different approach.2010-11-15