# General solution to $\sin \alpha + \sin \beta$ and $\cos \alpha + \cos \beta$?

For some angles $$\alpha,\beta$$, what is $$\sin\alpha+\sin\beta$$? What about $$\cos\alpha + \cos\beta$$?

My line of thought was to designate $$\theta=\alpha+\beta$$, for $$0\le\alpha\le 2\pi$$. By much experimentation, and scratching my head when I saw that $$\sin$$ needed a horizontal-shift term that depended on $$\theta$$ while $$\cos$$ didn’t, I eventually stumbled upon:

$$\sin\alpha + \sin\beta = \sin\alpha + \sin\left(\theta-\alpha\right) = \\ 2\sin\left(\frac{\theta}2\right)\sin\left(\alpha+\frac{\pi-\theta}2\right) = \\ 2\sin\left(\frac{\theta}2\right)\cos\left(\alpha-\frac{\theta}2\right)$$

and

$$\cos\alpha + \cos\beta = \cos\alpha + \cos\left(\theta-\alpha\right) = \\ 2\cos\left(\frac{\theta}2\right)\cos\left(\alpha-\frac{\theta}2\right)$$

That said, I’m not able to prove any of this. This was just from an afternoon of sketching out a bunch of test angles for $$\theta$$, plotting the curve $$\sin x + \sin(\theta-x)$$, and manipulating another function until it consistently matched the first. Is there a way to demonstrate numerically that the above equivalences are, in fact, correct?

• See, for instance, this question, for which I gave a geometric proof. I believe there are others versions of the question here, as well, that may more-directly address your request to demonstrate the result "numerically". – Blue Feb 13 '19 at 2:42
• FYI, there are general formulas for the sum & difference of $\sin$ and $\cos$, such as given in the Sum-to-product sub-section Wikipedia's Product-to-sum and sum-to-product identities. Your two results can be determined directly using these formulas. – John Omielan Feb 13 '19 at 3:03
• @JohnOmielan Yeah, I’m familiar with those formulae, but I just couldn’t figure out how to do it. Apparently I was just overthinking it. – DonielF Feb 13 '19 at 3:12

Using the sum/difference identities of the sine function, we find that $$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$$ $$\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta)-\sin(\beta)\cos(\alpha)$$ Adding them together, we find that $$\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin(\alpha)\cos(\beta)$$Now suppose that we want to find $$\sin(x) + \sin(y)$$ where $$x. Then we can rewrite $$x$$ and $$y$$ as $$\alpha \pm \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}, \quad \beta =y- \frac{x+y}{2}$$ $$x = \alpha - \beta, \quad y =\alpha + \beta$$ $$\sin(x) + \sin(y) = \sin(\alpha - \beta) + \sin(\alpha + \beta) = 2\sin(\alpha)\cos(\beta)$$