1
$\begingroup$

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 $\sin^2\alpha+\cos^2\alpha=1$?

$\endgroup$
  • 6
    $\begingroup$ What are your definitions of sine and cosine? This can be proved in several ways, which do not require appeal to geometric arguments, if your definition is analytic. (In terms of solutions to differential equations, or using power series, for example.) If your definition is geometric, you definitely need a geometric argument and, in that case, the "drawing" is just so you can follow the argument, it is not an essential part of it. $\endgroup$ – Andrés E. Caicedo Dec 15 '13 at 18:29
  • $\begingroup$ In total agreement with @AndresCaicedo, I saya that if your definition of sine is geometric, then you ought to look for a geometric proof; if your definition is the $y$-coordinate of the terminal ray’s intersection with the unit circle, you should look for a proof on the unit circle — it’ll be partly geometric, partly algebraic; if your definition comes out of the exponential function, your proof should be related to that; if your definition comes out of complex numbers, similarly. $\endgroup$ – Lubin Dec 15 '13 at 18:33
  • 1
    $\begingroup$ You will not be able to find a proof involving only the other trig identities. $\endgroup$ – Steven Gubkin Dec 15 '13 at 18:43
  • 1
    $\begingroup$ Well, the cosine satisfies the same kind of identity that the sine does, after all. Or: Let $S(x)=2x/(x^2+1)$ and $C(x)=(x^2-1)/(x^2+1)$. Then $S^2+C^2=1$, I think. $\endgroup$ – Lubin Dec 15 '13 at 18:53
  • 1
    $\begingroup$ Here's a relevant thread. I think the proof using rotation matrices is a nice non-picture proof. $\endgroup$ – littleO Dec 15 '13 at 18:59
4
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ This is the kind of proof I like best. $\endgroup$ – Lubin Dec 15 '13 at 18:57
  • $\begingroup$ Thanks. That's what i wanted. $\endgroup$ – CODE Dec 16 '13 at 11:20
  • $\begingroup$ @CODE : that's the proof french students get in high school, as an application of the scalar product. And then we use the result to introduce the exp notation of the complex numbers. $\endgroup$ – imj Dec 16 '13 at 18:30
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ كان هذا دليلا لطيفة ومتسقة. $\endgroup$ – user93957 Dec 15 '13 at 18:49
  • $\begingroup$ أين أنت من سامي؟ $\endgroup$ – user93957 Dec 15 '13 at 18:49
  • $\begingroup$ شكرا لك أدوب يسرني أن أراك تكتب باللغة العربية $\endgroup$ – user63181 Dec 15 '13 at 18:54
  • $\begingroup$ وذلك بفضل الترجمة من Google. :p $\endgroup$ – user93957 Dec 15 '13 at 18:55
  • $\begingroup$ Nicely done, @Sami! +1 $\endgroup$ – Namaste Dec 16 '13 at 14:27
3
$\begingroup$

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 $$ \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) $$

$\endgroup$
3
$\begingroup$

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 $$f(t)=\cos\beta\sin t+\sin\beta\cos t .$$


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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\cos(\alpha+\beta)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.