Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$

Question

Why is $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$? How can we find the RHS if we don't know what it is? (instead of proving the identity itself)
I could find a geometric solution in Wikipedia, but is there any solution that doesnt require drawing something?
Edit: I now saw some nice proofs using $e$ and Euler's identity. I would appreciate anything NOT using them too, as a change.
Out of curiosity, Is there a proof that says you actually "can't" prove the identity using only other simple 1-variable identities like $\sin^2\alpha+\cos^2\alpha=1$?

Answer 1

I usually like to reduce these to problems involving complex exponents. Note that

$$\sin\alpha\cos\beta + \sin\beta\cos\alpha = \frac{e^{i\alpha} - e^{-i\alpha}}{2i}\frac{e^{i\beta} + e^{-i\beta}}{2} + \frac{e^{i\beta} - e^{-i\beta}}{2i}\frac{e^{i\alpha} + e^{-i\alpha}}{2}$$ and simplify this expression.

Answer 2

Using the scalar product : let $A$ and $B$ be the points on the unit circle at arc length $a$ and $b$. Then $A(\cos(a);\sin(a))$ and $B(\cos(b);\sin(b))$, and $\widehat{BOA}=a-b$. Therefore :

\begin{aligned} \overrightarrow{OB}\cdot\overrightarrow{OA} &= OA\times OB\times \cos(\widehat{\overrightarrow{OB};\overrightarrow{OA}})\\ &= 1\times 1\times \cos(\widehat{BOA})\\ &= \cos(a-b) \end{aligned}

and as $\overrightarrow{OA}\binom{\cos(a)}{\sin(a)}$, $\overrightarrow{OB}\binom{\cos(b)}{\sin(b)}$ :

$$ \overrightarrow{OB}\cdot\overrightarrow{OA} = \cos(a)\cos(b)+\sin(a)\sin(b) $$

Therefore $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$

Then \begin{aligned} \sin(a-b)&=\cos\left(\frac{\pi}{2}+b-a\right)\\ &=\cos\left(\frac{\pi}{2}+b\right)\cos(a)+\sin\left(\frac{\pi}{2}+b\right)\sin(a)\\ &=-\sin(b)\cos(a)+\cos(b)\sin(a)\\ &= \sin(a)\cos(b)-\sin(b)\cos(a) \end{aligned}

Answer 3

To find the RHS (assuming that we may forget the result) we write:

$$\sin(\alpha+\beta)=\Im\frac{1}{2}(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)})\\=\Im\frac{1}{2}((\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)-(\cos\alpha-i\sin\alpha)(\cos\beta-i\sin\beta))$$ and we develop and we take the imaginary part we find the desired formula.

Answer 4

I like to see it using Euler's formula $e^{i\theta} = \cos{\theta} + i \sin{\theta}$. We have

$$ \begin{eqnarray} e^{i(\alpha+\beta)} = \cos(\alpha+\beta) + i \sin(\alpha+\beta) \end{eqnarray} $$

but also

$$ \begin{eqnarray} e^{i(\alpha+\beta)} &=& e^{i\alpha}\cdot e^{i\beta} \\\\ &=& [\cos(\alpha) + i \sin(\alpha)]\cdot [\cos(\beta) + i \sin(\beta)]\\\\ &=& \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) + i[\cos(\alpha)\sin(\beta) + \cos(\beta) \sin(\alpha)] \end{eqnarray} $$

So you get both $$ \sin(\alpha+\beta) = \cos(\alpha)\sin(\beta) + \cos(\beta) \sin(\alpha) $$ and $$ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) $$

Answer 5

The differential equation $$f''=-f$$ has a unique solution for given initial conditions $f(0)=x_0$, $f'(0)=y_0$. To show uniqueness assume that $g$ is another solution for the same initial values. Then $h:=f-g$ also satisfies $h''=-h$, and $h(0)=h'(0)=0$. But then $$\frac\partial{\partial t}(h(t)^2+h'(t)^2)=2h(t)h'(t)+2h'(t)h''(t)=2h(t)h'(t)-2h'(t)h(t)=0, $$ and it follows that $h(t)=0$ must hold for all $x$, so $f=g$.

This unique solution is $$f(t)=y_0\sin t+x_0\cos t, $$ as one checks immediately.

Now we can apply this to the function $f(t)=\sin(t+\beta)$. It satisfies $f''(t)=-f(t)$, $f(0)=\sin\beta$, $f'(0)=\cos\beta$, hence we must have $$f(t)=\cos\beta\sin t+\sin\beta\cos t .$$

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This may seem to come out of the blue at first, but if one sets $x(t)=f(t)$, $y(t)=f'(t)$, then $f''=-f$ becomes $$\begin{pmatrix}x\\y\end{pmatrix}'=\begin{pmatrix}y\\-x\end{pmatrix}, $$ which describes a motion on a circle around the origin (the velocity is perpendicular to the vector from the origin to the current point), and of course $\sin$ and $\cos$ are all about circles. The calculation done above for $h$ shows that indeed $x(t)^2+y(t)^2$ is constant.

Answer 6

If you don't want to use geometry, you can do it with complex powers of $e$: $$ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\\ \cos(x)=\frac{e^{ix}+e^{-ix}}{2}\\ \sin(x+y)=\frac{e^{i(x+y)}-e^{-i(x+y)}}{2i} $$ You can now also express the right hand side of your equality in terms of powers of $e$, and this wil give the same result.

Answer 7

Consider the vectors $$\vec{u}=(\sin \alpha,\cos \alpha)= \left(\cos \left(\frac{\pi}{2}-\alpha\right),\sin\left(\frac{\pi}{2}-\alpha\right)\right) $$

and $$\vec{v}=(\cos \beta,\sin \beta) $$ in $\mathbb R^2$. The dot product $\vec{u} \cdot \vec{v}$ can be calculated in two ways:

1. $\vec{u} \cdot \vec{v}=\|\vec{u}\| \|\vec{v}\| \cos \theta$ , where $\theta=\frac{\pi}{2}-\alpha-\beta$ is the angle between $\vec{u}$ and $\vec{v}$. Since $\vec{u},\vec{v}$ are unit vectors, this representation gives $\vec{u} \cdot \vec{v}=\cos \left( \frac{\pi}{2}-\alpha-\beta \right)=\sin(\alpha+\beta)$.

2. $\vec{u} \cdot \vec{v}=\sin \alpha \cos \beta +\sin \beta \cos \alpha$.

Compare the two.

Answer 8

Even though that is just the complex numbers argument in disguise...

You asked specifically how we can find the RHS. Now if you believe that rotations are linear maps and that a rotation by an angle of $\alpha$ followed by a rotation by an angle of $\beta$ is the same as a rotation by an angle of $\alpha+\beta$ then you are lead to \begin{align} D_{\alpha+\beta}&=D_\beta D_\alpha, & D_\phi&=\begin{pmatrix} \cos\phi&-\sin\phi\\ \sin\phi&\cos\phi \end{pmatrix}, \end{align} which is equivalent to the formula for $\sin(\alpha+\beta)$ and the corresponding one for $\cos(\alpha+\beta)$.