Trigonometry – Alternative Proof for Angle Difference Expansion

euclidean-geometrytrigonometry

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?

Best Answer

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 $\alpha+\beta$:

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$$\begin{align} \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \sin\alpha \cos \beta + \cos\alpha \sin \beta \\[6pt] |\cos(\alpha+\beta)| &= \sin\alpha \sin\beta - \cos\alpha \cos\beta \\ \to\qquad \phantom{|}\cos(\alpha+\beta)\phantom{|} &= \cos\alpha \cos\beta - \sin\alpha \sin\beta \end{align}$$


Non-Obtuse $\alpha$, with Obtuse $\beta$ and $\alpha+\beta \leq 180^\circ$:

enter image description here

$$\begin{align} \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \cos\alpha\sin\beta - \sin\alpha\,|\cos\beta| \\ \to\quad \phantom{|}\sin(\alpha+\beta)\phantom{|} &= \sin\alpha \cos\beta + \cos\alpha \sin\beta \\[6pt] |\cos(\alpha+\beta)| &= \cos\alpha\,|\cos\beta| + \sin\alpha \sin\beta \\ \to\quad \phantom{|}\cos(\alpha+\beta)\phantom{|} &= \cos\alpha\cos\beta - \sin\alpha\sin\beta \end{align}$$


The cases for $\alpha+\beta > 180^\circ$ are left as exercises to the reader.