I have thought about this for a while and have no progress. Does there exist a purely Euclidean Geometric proof of the Angle Difference expansion for Sine and Cosine, for Obtuse angles?
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You mean like this? – Blue Aug 03 '15 at 04:07
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No. I wish to prove the Obtuse case, which implies the acute case. Proving the acute case does not geometrically imply the obtuse case. – Jack Tiger Lam Aug 03 '15 at 04:15
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My diagrams can be adjusted to fit the obtuse case. Of course, one has to make the appropriate allowances for "negative" distances. – Blue Aug 03 '15 at 04:28
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My friend showed me a direct proof for the angle difference case, but it only holds true for the acute case. Inscribe a right angled triangle inside another one such that the right-angle vertex and another one coincide, with the remaining third vertex existing along the side of the major triangle. Label the hypotenuses and hence deduce the angle difference formula. – Jack Tiger Lam Aug 03 '15 at 06:54
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How do you figure the obtuse case geometrically implies the acute case but the acute case does not imply the obtuse case? If you are doing it with carefully labeled diagrams such as the ones in the answer below, which are basically just the acute-case diagrams with various relabelings of angles and sides (including "relabelings" that swap one side of an angle for a perpendicular ray), then wouldn't you have to relabel the figures all over again in order to prove the acute case? How does all this relabeling count as "implication" when you do it in one direction but not in the other? – David K Jan 23 '22 at 04:45
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As someone recently pointed out in another question, we never got to see how to relabel the figures to get difference formulas. But we don't actually even need the relabeling; based on just the two diagrams for the acute-case formulas and a few simple facts about the sine and cosine of angles outside the range $[0,\frac\pi2],$ we can get sum or difference formulas for any angles we want. I give a few examples here. – David K Jan 23 '22 at 04:51
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Here's a hint, in the form of adapting my Angle-Sum diagram to a couple of obtuse cases. Perhaps they'll guide you to adapting my Angle-Difference diagram appropriately.
Non-Obtuse $\alpha$ and $\beta$, with Obtuse $\alpha+\beta$:
$$\begin{align} \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \sin\alpha \cos \beta + \cos\alpha \sin \beta \\[6pt] |\cos(\alpha+\beta)| &= \sin\alpha \sin\beta - \cos\alpha \cos\beta \\ \to\qquad \phantom{|}\cos(\alpha+\beta)\phantom{|} &= \cos\alpha \cos\beta - \sin\alpha \sin\beta \end{align}$$
Non-Obtuse $\alpha$, with Obtuse $\beta$ and $\alpha+\beta \leq 180^\circ$:
$$\begin{align} \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \cos\alpha\sin\beta - \sin\alpha\,|\cos\beta| \\ \to\quad \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \sin\alpha \cos\beta + \cos\alpha \sin\beta \\[6pt] |\cos(\alpha+\beta)| &= \cos\alpha\,|\cos\beta| + \sin\alpha \sin\beta \\ \to\quad \phantom{|}\cos(\alpha+\beta)\phantom{|} &= \cos\alpha\cos\beta - \sin\alpha\sin\beta \end{align}$$
The cases for $\alpha+\beta > 180^\circ$ are left as exercises to the reader.
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I'm kind of stuck. How did you get sin(a+b)? When their sum is smaller than 90, we get it by saying that sin(a+b)= opp/1 and the opposite side will be sinacosb+ sinbcosa. But this time angle(a+b) isn't inside any triangle and I've been looking at the graph for more than 10 minutes to try to get how we arrived at it but in vain :( – Manar Jan 05 '21 at 22:16
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@Manar: Beyond $90^\circ$, the rules change; see, eg, this answer. What's not said there is that we can still associate the sin & cos of a larger angle $\theta$ w/ a right triangle: not one that contains $\theta$, but one built from $\theta$'s "reference angle" (an angle that brings $\theta$ to a multiple of $180^\circ$). In the figures above, the white triangle plays this role; the ref angle of $a+b$ is the one near the "m" in ".com". The "opp" & "adj" of that angle in that triangle (subject to a sign change or two), are the sin & cos of $a+b$. – Blue Jan 06 '21 at 16:41