How do you derive the formula of $\sin (a+b)$ from $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

I know that cos is even while sin is odd, and I know $\cos(\pi)=\sin((\pi/2)-x)$, but I still can't figure the derivation of $\sin (a+b)$ from $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$. Could you give me any hint?

• This can be proven easily from Euler's Formula. $e^{i \theta} = cos(\theta) + isin(\theta)$ – MadScientist Nov 14 '15 at 23:28
• Does it really derive the formula for sine from the formula for cosine? – Bernard Nov 14 '15 at 23:30
• @Bernard You could frame it that way, but I suppose it's sort of a cheat to do so. You could start by defining sin, cos and e^x all in terms of the taylor series. He's probably looking for a more elementary proof, but this one deserves to be mentioned because its so simple. – MadScientist Nov 14 '15 at 23:33

$$\sin(a+b)=\cos\left(\left(\frac{\pi}{2}-a\right)+(-b)\right)=\cos\left(\frac{\pi}{2}-a\right)\cos(-b)-\sin\left(\frac{\pi}{2}-a\right)\sin(-b)\\=\sin(a)\cos(-b)-\cos(a)(-\sin(b))=\sin(a)\cos(b)+\cos(a)\sin(b)$$

Since Element118 has already given what you expected, here is a geometric proof (the grey squares represent right angle).

By definition, we know that $$\sin(\alpha+\beta)=\frac{|BC|+|CD|}{|AB|}=\frac{|BC|}{|AB|}+\frac{|CD|}{|AB|}$$

We want $$\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)$$

We have $\cos(\alpha)=\frac{|BC|}{|BE|}$ and $\sin(\beta)=\frac{|BE|}{|AB|}$ and it gives: \begin{equation*} \frac{|BC|}{|AB|}=\frac{|BC|\cdot|BE|}{|AB|\cdot|BE|}=\underbrace{\frac{|BC|}{|BE|}}_{\cos(\alpha)}\underbrace{\frac{|BE|}{|AB|}}_{\sin(\beta)} \end{equation*} We repeat this for $\frac{|CD|}{|AB|}$ and we get

$$\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)$$

• @Blue Very interesting! Thanks. – MoebiusCorzer Nov 15 '15 at 1:01

A very short hint:

Deriving ;o)

and some details:

Start from $$\cos(a+b)=\cos a \cos b-\sin a\sin b$$ and differentiate w. r. t., say, $a$: $$-\sin(a+b)=-\sin a\cos b-\cos a \sin b.$$

• Is it so elliptic? – Bernard Nov 14 '15 at 23:36
• What is ;o) supposed to mean? – user223391 Nov 14 '15 at 23:46
• Bernard was just noting the humor that the OP was using "derive" in the colloquial sense of "figure out," but that coincidentally "take the derivative of" is a good way to do so. – Simpson17866 Nov 14 '15 at 23:47
• Sure, that piece of technical information was good to know and I'm surprised that I'd missed it for this long, but does the fact that a pun is not "technically" correct mean that the pun is "wrong"? If I said "I used to be a banker, but I lost interest," would the play on the word "interest" be a point in favor of the pun or a point against? – Simpson17866 Nov 14 '15 at 23:59
• In my opinion, it'd be a point in favour of the pun… – Bernard Nov 15 '15 at 0:01