Show $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ with the circle.

Question

How can I prove that $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ using the circle ?

Answer 1

Let $z_1=\cos a + i\sin a$ and $z_2=\cos b + i\sin b$ then $$ \eqalign{ z_1 \cdot z_2 &= (\cos a + i\sin a)\cdot(\cos b + i\sin b) \cr &= (\cos a \cos b - \sin a \sin b) + i(\cos a \sin b + \sin a \cos b) \cr } $$ we conclude $$ \cases{ \cos(a+b) = \cos a \cos b - \sin a \sin b \cr \sin(a+b) = \cos a \sin b + \sin a \cos b \cr } $$

Answer 2

I know that the OP is requesting a geometrical demonstration, but I thought that it would be instructive to see a couple of other approaches.

METHOD 1:

Let $f(x)$ be the function given by

$$f(x)=\cos x \cos(y-x)-\sin x \sin (y-x) \tag 1$$

Differentiating $(1)$, we find that

$$\begin{align} f'(x)&=-\sin x\cos (y-x)+\cos x \sin (y-x)-\cos x \sin (y-x)+\sin x \cos (y-x)\\\\ &=0 \end{align} $$

Therefore $f$ is a constant for all $x$.

First, letting $y=a+b$ and $x=0$ in $(1)$ shows that

$$f=\cos (a+b) \tag 2$$

Then, letting $y=a+b$ and $x=a$ in $(1)$ shows that

$$f=\cos a \cos b-\sin a \sin b \tag 3$$

Equating $(2)$ and $(3)$ reveals that

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

as was to be shown!

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METHOD 2:

Using Euler's Identity $e^{ix}=\cos x+i\sin x$, we have

$$e^{i(a+b)}=\cos (a+b)+i\sin(a+b) \tag 1$$

and

$$\begin{align} e^{i(a+b)}&=e^{ia}e^{ib}\\\\ &=(\cos a+i\sin a)(\cos b+i\sin b)\\\\ &=(\cos a \cos b-\sin a \sin b)+i(\sin a \cos b\cos a \sin b)\tag 2 \end{align}$$

Equating real and imaginary parts of $(1)$ and $(2)$, respectively, reveals

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

and

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

as expected!

Answer 3

I usually forget this kind of identities, so I have found that it is easy to get some of them writing sine and cosine as complex exponentials. In this case we have: $$\cos(a+b)=\frac{e^{i(a+b)}+e^{-i(a+b)}}{2}=\frac{e^{ia}e^{ib}+e^{-ia}e^{-ib}}{2}=$$ $$=\frac{(\cos a+i \sin a)(\cos b +i \sin b)+(\cos a - i \sin a)(\cos b - i \sin b)}{2}=$$ $$=\frac{\cos a \cos b + i \sin a \cos b + i \cos a \sin b -\sin a \sin b}{2}+\frac{\cos a \cos b - i \sin a \cos b - i \cos a \sin b -\sin a \sin b}{2}=$$ $$=\frac{2\cos a \cos b -2 \sin a \sin b }{2}= \cos {a} \cos {b} + \sin{ a} \sin {b}$$

PS: Sorry, I didn't read the part that said "using the circle", so this is not what you are looking for. Anyway, I leave it here in the case it is useful as an alternative method.

Answer 4

Rotate the point of coordinates $(\cos a, \sin a)$ around the origin with the rotation of angle $b$ using the matrix $$R(0,b)=\begin{pmatrix} \cos b & -\sin b\\ \sin b & \cos b \end{pmatrix}$$