All the proofs I've seen are geometrical, assuming that $A+B$ is less than $90$ degrees. How can you prove this identity for $A+B$ greater than $90$ degrees, or more generally, any arbitrary value?
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$\begingroup$ I'm sure you can just modify the usual arguments, even if the sum is greater than 90 - you'll just have to be careful with the signed lengths. $\endgroup$– Akiva WeinbergerCommented Aug 27, 2014 at 7:05
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1$\begingroup$ Easiest way, though, would probably be to mess with coordinates and the unit circle. $\endgroup$– Akiva WeinbergerCommented Aug 27, 2014 at 13:26
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$\begingroup$ See this recent answer. $\endgroup$– JimmyK4542Commented Aug 27, 2014 at 20:28
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2 Answers
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If you don't want a geometrical proof, then you need to indicate how you are defining $\cos$ and $\sin$. One way to define them is that $\cos(x)$ and $\sin(x)$ are the real and imaginary parts of $\exp(ix)$, and use the property that $$\exp(i(a+b)) = \exp(ia)\exp(ib)$$ Then $$\begin{align} \exp(i(a+b)) &= \exp(ia)\exp(ib)\\ &= (\cos(a) + i\sin(a))(\cos(b) + i\sin(b))\\ &= \cos(a)\cos(b) - \sin(a)\sin(b) + i(\sin(a)\cos(b) + \cos(a)\sin(b)) \\ \end{align}$$ And from this we obtain two trig identities at once: $$\cos(a+b) = \text{Re}(\exp(i(a+b))) = \cos(a)\cos(b) - \sin(a)\sin(b)$$ $$\sin(a+b) = \text{Im}(\exp(i(a+b))) = \sin(a)\cos(b) + \cos(a)\sin(b)$$
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$\begingroup$ But that requires knowing the Taylor series of sine, which requires knowing the derivative of sine... which involves knowing the summation formula. Besides, I think that this is much more heavy machinery than needed. I'm guessing that the OP wants a proof that is geometrical, but isn't restricted to $0\le\theta<90^\circ$. $\endgroup$ Commented Aug 27, 2014 at 6:59
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$\begingroup$ May be you could remove the reference to the power series that, in any manner, you don't use in your good answer. $\endgroup$ Commented Aug 27, 2014 at 7:42
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$\begingroup$ @ClaudeLeibovici But how else do you prove that sine and cosine are the real and imaginary parts of $\exp(ix)$? I know know way that doesn't involve differentiating the trig functions. $\endgroup$ Commented Aug 27, 2014 at 13:27
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$\begingroup$ @ClaudeLeibovici I just realized - you don't need to prove that $\exp(ix)$ is anything. You only need to prove De Moivre - that $\arg(wz)=\arg(w)+\arg(z)$. This can be done more easily, by noting that rotation is linear. $\endgroup$ Commented Aug 27, 2014 at 14:18
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$\begingroup$ I am glad to see that we totally agree ! Cheers :-) $\endgroup$ Commented Aug 27, 2014 at 19:02
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Hint:
Suppose that the two unit circles in the diagram are centered at the origin, with the second one rotated clockwise by $\measuredangle A$. Note that the two red line have equal length. Find the coordinates of $P,Q,R,\text{ and } S$ (but especially $Q$). Use this information to derive $\sin(A+B)$ and $\cos(A+B)$.
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$\begingroup$ Nice solution! Very elegant. $\endgroup$– user169852Commented Aug 27, 2014 at 21:21
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$\begingroup$ I can't find the method. Could you please explain? $\endgroup$ Commented Aug 29, 2014 at 10:09
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$\begingroup$ point $P$ has coordinates $(\cos A, \sin A)$. What are $Q,R,$ and $S$'s coordinates? $\endgroup$– John JoyCommented Aug 29, 2014 at 12:33
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