2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ This can be proven easily from Euler's Formula. $e^{i \theta} = cos(\theta) + isin(\theta)$ $\endgroup$ – MadScientist Nov 14 '15 at 23:28
  • $\begingroup$ Does it really derive the formula for sine from the formula for cosine? $\endgroup$ – Bernard Nov 14 '15 at 23:30
  • $\begingroup$ @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. $\endgroup$ – MadScientist Nov 14 '15 at 23:33
6
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

trig_form

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

\begin{equation} \sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha) \end{equation}

$\endgroup$
  • $\begingroup$ See also this answer. $\endgroup$ – Blue Nov 15 '15 at 0:54
  • 1
    $\begingroup$ @Blue Very interesting! Thanks. $\endgroup$ – MoebiusCorzer Nov 15 '15 at 1:01
4
$\begingroup$

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, $a$: $$-\sin(a+b)=-\sin a\cos b-\cos a \sin b.$$

$\endgroup$
  • $\begingroup$ Is it so elliptic? $\endgroup$ – Bernard Nov 14 '15 at 23:36
  • $\begingroup$ What is ;o) supposed to mean? $\endgroup$ – user223391 Nov 14 '15 at 23:46
  • $\begingroup$ 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. $\endgroup$ – Simpson17866 Nov 14 '15 at 23:47
  • 1
    $\begingroup$ 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? $\endgroup$ – Simpson17866 Nov 14 '15 at 23:59
  • 1
    $\begingroup$ In my opinion, it'd be a point in favour of the pun… $\endgroup$ – Bernard Nov 15 '15 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.