Friday, July 29, 2016

trigonometry - Converting a sum of trig functions into a product




Given,
$$\cos{\frac{x}{2}} +\sin{(3x)} + \sqrt{3}\left(\sin\frac{x}{2} + \cos{(3x)}\right)$$
How can we write this as a product?



Some things I have tried:




  • Grouping like arguments with each other. Wolfram Alpha gives $$\cos{\frac{x}{2}} + \sqrt{3}\sin{\frac{x}{2}} = 2\sin{\left(\frac{x}{2} + \frac{\pi}{6}\right)}$$but I don't know how to derive that myself or do a similar thing with the $3x$.

  • Write $3x$ as $6\frac{x}{2}$ and then using the triple and double angle formulas, but that is much too tedious and there has to be a more efficient way.

  • Rewriting $\sqrt{3}$ as $2\sin{\frac{\pi}{3}}$ and then expanding and trying to use the product-to-sum formulas, and then finally grouping like terms and then using the sum-to-product formulas, but that didn't work either.




I feel like I'm overthinking this, so any help or insights would be useful.


Answer



$$\cos{\frac{x}{2}} +\sin(3x) + \sqrt{3}\left(\sin\frac{x}{2} + \cos(3x)\right)$$



$$=\cos{\frac{x}{2}} + \sqrt{3}\sin\frac{x}{2} +\sin(3x) + \sqrt{3}\cos(3x)$$



$$=2\left(\frac{1}{2}\cos\frac{x}{2} + \frac{\sqrt{3}}{2}\sin\frac{x}{2} +\frac{1}{2}\sin(3x) + \frac{\sqrt{3}}{2}\cos(3x)\right)$$



Note that $\frac{1}{2}=\sin\frac{\pi}{6}$ and $\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$ so:




$$=2\left(\sin\frac{\pi}{6}\cos\frac{x}{2} + \cos\frac{\pi}{6}\sin\frac{x}{2} +\sin\frac{\pi}{6}\sin(3x) + \cos\frac{\pi}{6}\cos(3x)\right)$$



Then using Addition Theorem:



$$=2\left(\sin\left(\frac{x}{2}+\frac{\pi}{6}\right)+\cos\left(3x-\frac{\pi}{6}\right)\right)$$



$$=2\left(\sin\left(\frac{x}{2}+\frac{\pi}{6}\right)+\sin\left(3x+\frac{\pi}{3}\right)\right)$$



Then using Sums to Products:




$$=4\left(\sin\left(\frac{\frac{x}{2}+\frac{\pi}{6}+3x+\frac{\pi}{3}}{2}\right)\cos\left(\frac{\frac{x}{2}+\frac{\pi}{6}-3x-\frac{\pi}{3}}{2}\right)\right)$$



$$=4\sin\left(\frac{7x+\pi}{4}\right)\cos\left(\frac{-15x-\pi}{12}\right)$$


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