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The proofs that I've seen of the Sine Sum Formula $ \sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a) $ are from Khan Academy and Socratic. Both of them begin with geometric constructions like this:

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My question is how can you prove this geometric construction is allowed through Rigid Transformations. I don't see it as given that the hypotenuse can be 1, for any triangle created by angle A or created by angle B. I know that it's True, I just want to understand why we are arbitrarily able to stack right triangles like this and pick a hypotenus length, using Rigid Transformations (aka compass and straightedge).

Ben G
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  • Even for the particular case of sine and cosine of 20 degrees, there is no ruler and compass construction for those two lengths. [In general a sine or cosine of $n$ degrees, $n$ a positive integer between 0 and 360, can be constructed with ruler and compass if and only if $n$ is a multiple of $3.$] The proof is assuming you have those lengths, and proceeding to show the sum formula. It isn't a question of restricting to ruler and compass constructions. – coffeemath Nov 12 '21 at 19:58
  • Why are you allowed to the assume those lengths can exist? The sine sum formula is supposed to be used on any a and b, shouldn't its only assumptions be axioms? – Ben G Nov 12 '21 at 20:01
  • Well it isn't much of an assumption that for any specific positive real number there is a geometric segment of that length. Once the real numbers have been introduced and developed, using the axioms of a complete ordered field (as done in any good real analysis text), there is enough foundation even to deal with a segment of length $\pi,$ say. – coffeemath Nov 12 '21 at 20:06
  • So far I just have Geometry under my belt and was hoping that they'd prove that assumption using geometry. I was also just reading on wiki that.. rotation matrices may be an easier way to understand this? – Ben G Nov 12 '21 at 20:15
  • Also just found this video https://youtu.be/In6NZCp4cNA which creates a bigger triangle by placing them both on the unit circle and rotating. – Ben G Nov 12 '21 at 20:21
  • Strictly speaking, this is a proof in the case when $\alpha$ and $\beta$ are acute angles. If you want a proof for arbitrary $\alpha$ and $\beta$, you have to use something else. That said, this is a valid proof when $\alpha$ and $\beta$ are acute. So are you trying to understand why this proof is OK in the specific case of acute angles, or are you just looking for a more general proof where the angles are arbitrary? – bjorn93 Nov 12 '21 at 20:21
  • I'm fine with the proof for just acute angles. Also, just saw this question which showed a novel approach for non-acute https://math.stackexchange.com/questions/1108447/proving-sine-of-sum-identity-for-all-angles – Ben G Nov 12 '21 at 20:21
  • The homothety, a simple geometric construction, allows you to rescale the triangles the way you want. –  Nov 12 '21 at 21:17
  • For a calculus-based proof, $e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}$, which can be expanded as $\cos(\alpha+\beta)+i\sin(\alpha+\beta)=(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))+i(\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta))$ by means of Euler's formula. –  Nov 12 '21 at 21:22

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For a given $c>0$ and acute angle $\theta$ (where $\theta$ is arbitrary in the interval $(0,90^{\circ}))$, you can construct a right triangle with hypotenuse $c$ and angle $\theta$. If you want to use rigid transformations, the construction goes like this. Draw $AC=c$, then rotate $C$ by $\theta$ degrees, and let its image be $C_1$. Reflect $C$ about the line $AC_1$, and let the reflection be $C_2$. Let lines $AC_1$ and $CC_2$ meet at $B$. Then your right triangle is $\triangle ABC$ with $\angle B=90^{\circ}$ and $\angle A=\theta$.

The proof you cited then could say something like this. Draw a right triangle with hypotenuse $1$ and acute angle $\beta$. Its legs are $\cos\beta$ and $\sin\beta$. Draw another right triangle with hypotenuse $\cos\beta$ (i.e. you use the leg of first triangle as a hypotenuse) and acute angle $\alpha$. That's how you arrive at the picture. The choice of $1$ for the length of the hypotenuse is irrelevant. You could choose any $c>0$. It's just that $c=1$ simplifies computations.

bjorn93
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  • Okay, I think I got it now. You start by showing that you can produce arbitrarily sided right triangles (size 1 is good for simplicity). Then you construct an arbitrarily sided acute right triangle with a hypotenuse equal to the leg of the previous one and stack it to form a bigger angle. – Ben G Nov 12 '21 at 22:41
  • @bgcode Right but just note that a right triangle is a right triangle, and an acute triangle is an acute triangle, can't be both at the same time. A right triangle has two acute angles though. – bjorn93 Nov 12 '21 at 23:40