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Rewrite the following expression as a simplified expression containing one term: $$\cos (\frac{\pi}{3}+\varphi) \cos (\frac{\pi}{3}-\varphi) - \sin (\frac{\pi}{3}+\varphi) \sin (\frac{\pi}{3}-\varphi)$$

$$\cos [(\frac{\pi}{3}+\varphi)+(\frac{\pi}{3}-\varphi)]= \cos (\frac{2\pi}{3})$$

Correct, or no? Thanks

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Recall that $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$

Yes $$\cos((\frac{\pi}{3}+\phi)+(\frac{\pi}{3}-\phi))=\cos(\frac{2\pi}{3})$$

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The best way to handle such expressions is probably the Euler's formula :

$cos(φ) = \dfrac{e^{jφ}+e^{-jφ}}{2}$, $sin(φ) = \dfrac{e^{jφ}-e^{-jφ}}{2}$

Very complex expressions can be tough to handle with trigonometry

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