Is there a visual proof of the addition formula for $\sin^2(a+b)$ ? The visual proof of the addition formula for $\sin(a+b)$ is here :

enter image description here

Also it is easy to generalize (in any way: algebra , picture etc) to an addition formula for $\sin^n(a+b)$ where $n$ is a given positive integer ?



$1)$ I prefer the addition formula's to have as little sums as possible. I assume this is equivalent to allowing and preferring large power of $\sin$ and $\cos$ ; e.g. $\sin^4(a+b)=$ expression involving $\sin^2$, $\sin^4$ and $\cos^4$ and no other powers of $\sin$ or $\cos$.

In one of the answers, the poster just used the binomium. That works nice, but Im not sure I like the output of that answer. We get $n$ sums for $sin^n(a+b)$ and I assume we can do better if we allow powers of $\sin$ and $\cos$.

I could be wrong ofcourse.

$2)$ I bet against the existance of visual proofs for $\sin^n(a+b)$ for $n>1$. But I could be wrong.

$3)$ Not trying to insult the answers and comments but I am skeptical about the use of complex numbers (and $\exp$) to solve this issue EFFICIENTLY. I know Euler's formula for $\exp(i x)$ but still. I could be wrong about this too ofcourse.

$4)$ My main intrest is in the $\sin^2$ case. I assume it has many forms. Can the addition formula for $\sin^2$ be expressed by $sin^2$ only ? I think so. (One of the reason I think so is because $\cos^2$ can be rewritten.)

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    $\begingroup$ Try to think of $\sin^n(a+b)$ as of ${\rm Im} \left[e^{i n(a+b)}\right]$ and then geometry on complex plane will became easy. $\endgroup$ – 0x2207 Jul 14 '13 at 14:09
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    $\begingroup$ @0x2207 Yeah, I tried to do that, too, but first of all taking the imaginary part so directly gives you the wrong stuff (try to write it down, you'll see), and then if you do it right you get to a point where you have an ugly induction to do. Using the binomial theorem is much easier. $\endgroup$ – Daniel Robert-Nicoud Jul 14 '13 at 14:30
  • $\begingroup$ See my edit guys. Thanks for the comments. $\endgroup$ – mick Jul 14 '13 at 14:56
  • $\begingroup$ I hope I am not trivializing this, but if you are looking for a "visual proof", have you considered making a 3D box, using the image you provided for two adjacent sides, connected along $ \cos \alpha \cos \beta $ side of each. Then you see that foiling out the $ ( \cos \alpha \sin \beta + \sin \alpha \cos \beta )^2 $ gives you an identity. This does not really address your desire to have a form with only sine in it though. Also, it might be messy to draw well. $\endgroup$ – N. Owad Jul 14 '13 at 15:41
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    $\begingroup$ @DanielRobert-Nicoud, yes, you are right, my formula is wrong. $\endgroup$ – 0x2207 Jul 15 '13 at 7:52

It's not entirely clear what you mean by "the addition formula for $\sin^2(\alpha+\beta)$", but if it's this ...

$$\sin^2(\alpha+\beta) = \cos^2 \alpha + \cos^2\beta - 2 \cos\alpha\cos\beta \cos\left(\alpha+\beta\right)$$

... then here's a picture-proof that relies on the Law of Cosines (which itself has a nice picture proof).


We simply inscribe $\alpha$ and $\beta$ to either side of a unit-length diameter of a circle, and apply the Law to the green-red-blue cos-cos-sin triangle. (The dashed right(!) triangle (re-)confirms why the blue segment has length $\sin(\alpha+\beta)$.)

Note: The figure also illustrates Ptolemy's Theorem ---The product of the diagonals of an inscribed quadrilateral is equal to the sum of the products of opposite sides--- since the unmarked green and red edges have lengths $\sin\alpha$ and $\sin\beta$, respectively, so that $$1 \cdot \sin(\alpha+\beta) = \sin\alpha \cos\beta + \sin\beta \cos\alpha$$

Note 2: The figure also gives this version of the addition formula ...

$$\sin^2\left(\alpha+\beta\right) = \sin^2\alpha + \sin^2\beta + 2 \sin\alpha\sin\beta\cos\left( \alpha+\beta \right)$$ once we interpret the right-hand side as $\sin^2\alpha + \sin^2\beta - 2 \sin\alpha\sin\beta\cos\left( \left(\frac{\pi}{2}-\alpha\right)+\left(\frac{\pi}{2}-\beta\right) \right)$ and apply the Law of Cosines to the green-red-blue sin-sin-sin triangle at the top of the picture. That's less pretty, though.

  • $\begingroup$ Thanks. I wonder if it can be done from law of sines. $\endgroup$ – mick Jul 15 '13 at 22:36
  • $\begingroup$ @mick: If you're seeking specific avenues of proof, then you should probably clarify exactly what you think the formula is. $\endgroup$ – Blue Jul 15 '13 at 22:48

For your second question:

We know that $\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)$. Then: $$\sin^n(a+b) = (\sin(a)\cos(b)+\cos(a)\sin(b))^n=$$ $$=\sum_{k=0}^n\binom{n}{k}\sin^k(a)\cos^k(b)\cos^{n-k}(a)\sin^{n-k}(b)$$ where we have used the binomial theorem for the last equality.

  • $\begingroup$ Thanks. For remarks on that, see my OP EDIT. $\endgroup$ – mick Jul 14 '13 at 14:56

A partial answer to the comments. Using half-angle for $\sin$, double angle for $\sin$ (or $\cos$) and $\sin^2 + \cos^2=1$ and the binomial theorem repeatedly we can reduce all expressions to $\sin^2$ forms.

(We Always take the binomial theorem first, and usually half-angle second and then continu with half and double in any practical order)

We could also use this method -together with rewriting $\sin^a$ into lower powers $(b<a)$ - to rewrite any addition formula with $m<n+1$ into powers of $m$. Thus expression involving only $\sin^3$ or only $\sin^4$ etc.

This answers comment $1)$ partially and comment $4)$ completely, although it must be said that the method may be very far from optimal. Seems like a brute force algoritm, but at least it works.

After reading the comments and answers of others, it appears that today(15/07/13) the main questions are: 1) Can all this be done more efficient. 2) How about those visual proofs ??

I am reminded of a similar question: Can we visually prove the addition formula for $\sin(a+b+c)$ without giving the proof for - or using - the addition formula for $\sin(a+b)$ ?

I assume that is not related but again, I could be wrong.

Not a complete answer, just my 50 cents.

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    $\begingroup$ It's a separate question, but yes ... You can visually prove the addition formula for $\sin(a+b+c)$. Just build a figure like the one in your question above, but with angles $a$, $b$, $c$ meeting at a point; draw in a zillion right triangles, and the result falls (right!) out. $\endgroup$ – Blue Jul 15 '13 at 23:02

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