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This question already has an answer here:

I don't know if I'm asking for too much, but the proofs I've seen of the statement

$$\sin(x+y) =\sin(x)\cos(y) + \cos(x)\sin(y)$$

consist of drawing a couple of triangles, one on top of each other and then figuring out some angles and lengths until they arrive at the identity.

And I agree with the proof, is just that, even by flipping the triangle around, it only proves the identity for the case $x+y<\pi/2$, or if it does prove it for all values of $x$ and $y$, I wouldn't understand why.

As to construing a proof by using Euler's identity or the derivatives of sin and cos, I would ask the writer to first prove his/her already accepted formulas without using the addition identity.

So that is my humble question. How could one prove that for all the values of $x$ and $y$, the identity $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ holds.

Any thoughts/ideas would be really appreciated.

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marked as duplicate by Trevor Gunn, Sahiba Arora, Daniel W. Farlow, Namaste, Leucippus Aug 6 '17 at 1:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/questions/635077/… $\endgroup$ – JohnColtraneisJC Aug 5 '17 at 23:47
  • $\begingroup$ If the couple-of-triangles figure you mention is mine, then it happens that you can adjust the figure to fit obtuse cases. $\endgroup$ – Blue Aug 6 '17 at 0:05
  • $\begingroup$ @Blue. Could You explain how? For instance, what if $α$ is $20$ degrees and $β$ is $140$ degrees? $\endgroup$ – Leo Aug 6 '17 at 0:12
  • $\begingroup$ @Leo: That case (non-obtuse $\alpha$ and obtuse $\beta$) is covered by the second image in the answer I referenced. $\endgroup$ – Blue Aug 6 '17 at 0:23
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    $\begingroup$ This is NOT a duplicate of the question pointed out by Benjamin Moss. If the two questions are read CAREFULLY, taking all of several seconds to do that, it will be seen that there is a substantial difference. This question takes it to be already established that the identity is proved in the case where the sum is less than a right angle. That means the already established case can actually be used as a lemma in the proof of the general case. $\endgroup$ – Michael Hardy Aug 6 '17 at 1:25
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My favorite proof is based on transformation matrices. If you want to rotate a point $(x,y)$ counter-clockwise around the origin by $t$ radians, you can use matrix multiplication:

$$\begin{bmatrix}\cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$$

The product will be the coordinates of the newly rotated point.

So, suppose you want to rotate $(x,y)$ by $a + b$ radians. You could either do this in one go, or you could first rotate by $a$ radians and then by $b$. Either way, of course, you should end up with the same point. In other words,

$$\begin{bmatrix}\cos (a+b) & -\sin (a+b) \\ \sin (a+b) & \cos (a+b) \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$$ and $$\begin{bmatrix}\cos b & -\sin b \\ \sin b & \cos b \end{bmatrix}\begin{bmatrix}\cos a & -\sin a \\ \sin a & \cos a \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}\cos a \cos b - \sin a \sin b & -\sin a \cos b - \cos a \sin b\\ \cos a \sin b + \sin a \cos b & -\sin a \sin b + \cos a \cos b\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$$

should give us the same products. This requires that the two matrices must be equal, so $\sin(a+b) = \cos a \sin b + \sin a \cos b$ (and also, $\cos(a+b) = \cos a \cos b - \sin a \sin b$!).

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As a reference, consider page $125$ of these notes. By defining the complex exponential function through the everywhere-convergent series $$e^z\stackrel{\text{def}}{=}\sum_{n\geq 0}\frac{z^n}{n!} $$ it is simple to check that $e^{z}\cdot e^{w} = e^{z+w}$: $$ e^{z+w}=\sum_{n\geq 0}\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}z^k w^{n-k}=\sum_{a,b\geq 0}\frac{z^a}{a!}\cdot\frac{w^b}{b!}=e^{z}\cdot e^{w}. $$ In particular, for any $\theta\in\mathbb{R}$ the squared modulus of $e^{i\theta}$, given by $e^{i\theta}\cdot\overline{e^{i\theta}} = e^{i\theta}\cdot e^{-i\theta}$, equals $1$. Since $\frac{d}{dz}e^z=e^z$ is a trivial consequence of the series definition, we have that the map $\theta\to e^{i\theta}$ is an arc-length parametrization of the unit circle and the functions defined by $$\sin\theta = \text{Im }e^{i\theta},\qquad \cos\theta=\text{Re }e^{i\theta}$$ are the elementary trigonometric functions we already know and the statement $\left\|e^{i\theta}\right\|^2=1$ is equivalent to the Pythagorean Theorem. The interesting consequence is that:

$$\begin{eqnarray*}\sin(\theta+\varphi) = \text{Im}\left(e^{i\theta}\cdot e^{i\varphi}\right)&=&\text{Im}\left[\left(\cos\theta+i\sin\theta\right)\cdot\left(\cos\varphi+i\sin\varphi\right)\right]\\&=&\sin\theta\cos\varphi+\sin\varphi\cos\theta.\end{eqnarray*} $$


As an alternative approach, one may notice that for a fixed value of $\varphi$ both $f(\theta)=\sin(\theta+\varphi)$ and $g(\theta)=\sin\theta\cos\varphi+\cos\theta\sin\varphi$ are solutions of the differential equation

$$ f''+f=0,\quad f(0)=\sin\varphi,\quad f'(0)=\cos\varphi $$

hence they are the same function by the unicity part of the Cauchy-Lipschitz Theorem.

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