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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.

enter image description here

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.

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  • $\begingroup$ What are $\vec v \text{ and }\vec w?$ $\endgroup$ – saulspatz Mar 11 '18 at 0:44
  • $\begingroup$ I know a proof that uses inner product and fact that $\sin(\pi/2 - a) = \cos a$ and $\cos(\pi/2 - a) = \sin a$. $\endgroup$ – Mathsource Mar 11 '18 at 1:00
  • $\begingroup$ @MathFacts. That sounds interesting, do you have a link? $\endgroup$ – Leo Mar 11 '18 at 1:02
  • $\begingroup$ @saulpatz. I'm having trouble explaining what $v⃗$ and $w⃗$ are in words, so I apologze if my next explanation isn't very good. Imagine that we turn the circle $a$ degrees clockwise, then, $v⃗$ would be pointing straight up, while $w⃗$ would be pointing straight to the left, also their magnitudes are defined as $1+cos(b)$ and $sin(a)$ respectively, so that $v⃗+w⃗$ is the vector which starts at the origin and points to the dot in the circle whose respective angle is $b$. $\endgroup$ – Leo Mar 11 '18 at 1:08
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I get $\|w\| = 1 - \cos \beta$, not $\|w\| = \cos \beta$.

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  • $\begingroup$ You are right! However, I still don't get that $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$ but rather that $sin(a+b)=(-1)sin(a)cos(b)+cos(a)sin(b)$ $\endgroup$ – Leo Mar 11 '18 at 0:59
  • $\begingroup$ Ah, whoops, you're right. It should really be $\|w \| = 1 + \cos(\beta - 90^\circ) = 1 - \cos \beta$. That should fix it. $\endgroup$ – Michael Biro Mar 11 '18 at 1:13
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@Michael's answer identifies the error, but I thought I'd provide a comprehensive diagram. (It's kinda what I do.)

enter image description here

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.

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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 $$ \begin{split} \sin(a + b) &= \cos[\pi/2 - (a + b)] = \cos(\pi/2 - a)\cos b + \sin(\pi/2 - a)\sin b\\ &= \sin a\cos b + \sin b\cos a \end{split} $$

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