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?
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$\begingroup$ This can be proven easily from Euler's Formula. $e^{i \theta} = cos(\theta) + isin(\theta)$ $\endgroup$– MadScientistNov 14, 2015 at 23:28
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$\begingroup$ Does it really derive the formula for sine from the formula for cosine? $\endgroup$– BernardNov 14, 2015 at 23:30
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$\begingroup$ @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. $\endgroup$– MadScientistNov 14, 2015 at 23:33
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3 Answers
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$$\sin(a+b)=\cos\left(\left(\frac{\pi}{2}-a\right)+(-b)\right)=\cos\left(\frac{\pi}{2}-a\right)\cos(-b)-\sin\left(\frac{\pi}{2}-a\right)\sin(-b)\\=\sin(a)\cos(-b)-\cos(a)(-\sin(b))=\sin(a)\cos(b)+\cos(a)\sin(b)$$
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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
\begin{equation} \sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha) \end{equation}
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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, $a$: $$-\sin(a+b)=-\sin a\cos b-\cos a \sin b.$$
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$\begingroup$ What is ;o) supposed to mean? $\endgroup$– user223391Nov 14, 2015 at 23:46
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$\begingroup$ 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. $\endgroup$ Nov 14, 2015 at 23:47
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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? $\endgroup$ Nov 14, 2015 at 23:59
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1$\begingroup$ In my opinion, it'd be a point in favour of the pun… $\endgroup$– BernardNov 15, 2015 at 0:01