Adapting a picture proof of trig's Angle Addition identities to angles greater than $90^\circ$

I am currently learning about the Trig addition formula. The book I am reading has a proof for it. The proof is attached here:

However, I have a question about it. What if the values of theta and phi were greater than 90 degrees, or basically so great that you can't fit them in a right triangle. Since the proof relies on right triangles, would it still be legitimate in such a scenario? Can someone please explain? Don't give me another proof, simply explain if the proof in the book is legitimate for the values of theta and phi were greater than 90 degrees. And if not, then give me another proof.

Can you please explain as simply as possible. I am a high school student still learning pre-calculus, without knowledge of rigorous proofs. Can you try to keep the answer simple, and show and explain all your working so that I can understand?

• This answer of mine adapts a previous answer (showing my own version of the picture proof of the angle-sum identity) to obtuse angles. – Blue Jul 17 '18 at 10:50
• I suppose one could give a proof using rotations of vectors that works for any angles, even angles greater than $\pi,$ without needing to be concerned about how to fit the angles into right triangles. But rotations aren't usually taught until long after the course in trigonometry, so I suppose such a proof wouldn't help answer this question. – David K Jul 17 '18 at 11:53
• What is your definition of $\cos$ and $\sin$ for angles outside the interval $\bigl[0,{\pi\over2}\bigr]\,$? – Christian Blatter Jul 17 '18 at 11:57

You are convinced with the picture about the case where $$\theta +\phi < \pi/2$$ Then in case of $$\theta +\phi > \pi/2$$ we have $$\pi - (\theta +\phi) < \pi/2$$

$$\sin(\theta +\phi)=\sin (\pi -(\theta +\phi)) = \sin ((\pi /2- \theta)+(\pi /2- \phi))$$

Note that the new angles $$(\pi /2- \theta) \text { and }(\pi /2-\phi)$$ fit the picture and you get the desired result.

If all you want to do is to handle the case where $90^\circ < \theta + \phi < 180^\circ,$ then the figure looks like this:

You need to know how $\sin(\theta + \phi)$ is defined when $\theta + \phi > 90^\circ.$ Once you know the definition, it will be clear that $\sin(\theta + \phi) = \frac{PQ}{OP},$ and the rest of the proof proceeds as before.

Some material to digest:

When calculus is not avaiable the definition of $\cos$ and $\sin$ is usually extended to all of ${\mathbb R}$ in the following way: Given any $\phi\geq0$ wind a string of length $\phi$ counterclockwise around the unit circle $C$ of the $(x,y)$-plane, beginning at $(1,0)$. The end of the string will then be at a certain point $(x_\phi,y_\phi)\in C$. If $\phi<0$ do the same thing clockwise with a string of length $|\phi|$. In both cases define $$\cos\phi:=x_\phi,\quad\sin\phi:=y_\phi\ .\tag{1}$$ Note that for $0\leq\phi\leq{\pi\over2}$ this coincides with the "old" definition in terms of right triangles. So far we have extended the definition of $\cos$ and $\sin$. We now have to extend the laws, e.g. the addition formulas. To this end we make use of the following fact of analytic geometry: The map $$T:\quad{\mathbb R}^2\to{\mathbb R}^2,\qquad (x,y)\mapsto(-y,x)$$ rotates all points ${\pi\over2}$ counterclockwise around the origin. This allows us to introduce the following trick: For any $\phi\in{\mathbb R}$ write $$J\phi:=\phi+{\pi\over2}\ ,$$ and iterate this to $$J^n\phi:=\phi+n{\pi\over2}\qquad(n\in{\mathbb Z})\ .$$ We then may write $$\bigl(x_{J\phi},y_{J\phi}\bigr)=\bigl(x_{\phi+\pi/2},y_{\phi+\pi/2}\bigr)=T(x_\phi,y_\phi)=(-y_\phi,x_\phi)\ .\tag{2}$$ Using the definition $(1)$ of $\cos$ and $\sin$ we can now read off from $(2)$ the identities $$\cos\left(\phi+{\pi\over2}\right)=-\sin\phi,\qquad \sin\left(\phi+{\pi\over2}\right)=\cos\phi\ ,\tag{3}$$ valid for all $\phi\in{\mathbb R}$. Formula $(3)$ and its iterates then allow to "transport" the addition theorems, and similar laws, to all of ${\mathbb R}$.