 # General solution to sin &#x2061;<!-- ⁡ --> rjawbreakerca 2022-07-07 Answered
General solution to $\mathrm{sin}\alpha +\mathrm{sin}\beta$ and $\mathrm{cos}\alpha +\mathrm{cos}\beta$?
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Using the sum/difference identities of the sine function, we find that
$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)+\mathrm{sin}\left(\beta \right)\mathrm{cos}\left(\alpha \right)$
$\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)-\mathrm{sin}\left(\beta \right)\mathrm{cos}\left(\alpha \right)$
Adding them together, we find that
$\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)=2\mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)$
Now suppose that we want to find $\mathrm{sin}\left(x\right)+\mathrm{sin}\left(y\right)$ where x<y. Then we can rewrite x and y as $\alpha ±\beta$ where $\alpha$ is the average of x and y and where $\beta$ is the difference between y and the average. From there, we can plug it into the identity above. A similar computation can be arrived at using the sum/difference formulas for cosine.
$\alpha =\frac{x+y}{2},\phantom{\rule{1em}{0ex}}\beta =y-\frac{x+y}{2}$
$x=\alpha -\beta ,\phantom{\rule{1em}{0ex}}y=\alpha +\beta$
$\mathrm{sin}\left(x\right)+\mathrm{sin}\left(y\right)=\mathrm{sin}\left(\alpha -\beta \right)+\mathrm{sin}\left(\alpha +\beta \right)=2\mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)$

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