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?
Best Answer
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 $x<y$. 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. $$\alpha = \frac{x+y}{2}, \quad \beta =y- \frac{x+y}{2}$$ $$x = \alpha - \beta, \quad y =\alpha + \beta$$ $$\sin(x) + \sin(y) = \sin(\alpha - \beta) + \sin(\alpha + \beta) = 2\sin(\alpha)\cos(\beta)$$