Mistake in proof of $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

Question

I'm trying to prove the identity $$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ by using a specific vector and scalar method.

However, there has to be a mistake somewhere, since the equation at the end isn't correct. Such mistake I can't find.

I would really appreciate any help/thoughts.

Answer 1

I get $\|w\| = 1 - \cos \beta$, not $\|w\| = \cos \beta$.

Answer 2

Let $\vec{u} = (\cos a, \sin a)$ and $\vec{v} = (\cos b, \sin b)$. $$ \vec{u}\cdot \vec{v} = \cos a \cos b + \sin a \sin b $$ But, $$ \vec{u}\cdot \vec{v} = |\vec{u}||\vec{v}|\cos(b - a) = 1.1.\cos(b - a) = \cos(b - a) $$ Thus, $\cos(b - a) = \cos a \cos b + \sin a \sin b$ and consequently $$ \begin{split} \sin(a + b) &= \cos[\pi/2 - (a + b)] = \cos(\pi/2 - a)\cos b + \sin(\pi/2 - a)\sin b\\ &= \sin a\cos b + \sin b\cos a \end{split} $$

Answer 3

@Michael's answer identifies the error, but I thought I'd provide a comprehensive diagram. (It's kinda what I do.)

This shows

$$\begin{align} \sin(\alpha + \beta) &= \sin\alpha + \cos\alpha \sin\beta - \sin\alpha(1-\cos\beta) = \phantom{-\left(\;\right.}\sin\alpha \cos\beta + \cos\alpha \sin\beta \phantom{\left.\;\right)} \\ -\cos(\alpha+\beta) &= \cos\alpha(1-\cos\beta) + \sin\alpha \sin\beta - \cos\alpha = -\left(\;\cos\alpha\cos\beta - \sin\alpha\sin\beta\right)\; \end{align}$$

Note: The figure is valid even for non-obtuse $\beta$ (or non-acute $\alpha$), provided appropriate consideration is given to signed lengths.

Note 2: OP (and others) may be interested in a related figure of my own.