Visual proof of the addition formula for $\sin^2(a+b)$? 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 : 

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 ?
EDIT
4 COMMENTS :
$1)$ 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.)

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.
