# Sine of the sum of angles

I need to prove the formula for the sine of the sum $$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$ I already know how to prove it when $\alpha,\beta\geq 0$ and $\alpha+\beta <\pi/2$. How can I extend it to any pair of angles? The definition of sine and cosine that I am using is the length of the $y$-axis and the $x$-axis respectively when you intersect the circle of radius 1, but I can't use analytic geometry. Also I can't use complex numbers multiplication. Only relations like $\sin(\alpha +\pi/2) = \cos(\alpha)$.

• This definition is a dead end. You have to take by definition $(\cos \alpha, \sin \alpha)$ are the coordinates of point $M$ on the trigonometric circle (centered in $O$, radius 1) where angle $\alpha$ = angle AOM where A has coordinates $(1,0)$. In this framework, the formula expresses a dot product. – Jean Marie Apr 10 '16 at 22:52
• @Shailesh I can't use complex numbers – Minion Apr 10 '16 at 22:54
• @Shailesh in your formulas you have forgotten the $i$ in $e^{ix}$ etc... – Jean Marie Apr 10 '16 at 22:55
• @Shailesh last thing, the plus sign should be a minus sign in the definition of $\sin x$ – Jean Marie Apr 10 '16 at 22:57
• @JeanMarie. Thanks. I got rid of the comment. Too early in the morning for me. – Shailesh Apr 10 '16 at 23:00

There is a way, but is quite messy. You should first prove geometricaly that the formula is true for angles $-\pi/2 < \alpha,\beta < \pi/2$ such that $0\leq\alpha + \beta <\pi/2$. This can be done using the same construction you must have done for positive angles.
Now that you know that, suppose that $\pi/2\leq \alpha + \beta <\pi$. Then you can use that $\sin(\alpha+\beta) = \cos(\alpha+\beta -\pi/2)$ and use the formula of the cosine of the sum to obtain $$\cos((\alpha-\pi/2)+\beta) = \cos(\alpha-\pi/2)\cos(\beta) - \sin(\alpha-\pi/2)\sin(\beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\alpha)$$