How to show that $\langle \textbf {a,b} \rangle = \|\textbf {a}\| \|\textbf {b}\| \cos \theta$ where $\langle \cdot , \cdot \rangle$ denotes the usual real inner product or dot product on $\Bbb R^2$, $\| \cdot \|$ is the usual or Euclidean norm on $\Bbb R^2$ and $\theta$ is the angle between the vectors $\textbf {a}$ and $\textbf {b}$ in $\Bbb R^2$.
I have proved this result when $\theta$ is acute. How do I prove this result for obtuse or reflex angles? I didn't find any proof in the online sources which prove the above fact for obtuse or reflex angles. Would somebody help me in this regard?
Thank you very much.
EDIT $:$
Assume that $\theta$ is acute. WLOG by proper choice of coordinate axes we may assume that both $\textbf {a}$ and $\textbf {b}$ lie on the first quadrant. Let $\textbf {a} = (x,y)$ and $\textbf {b} = (z,w)$. Let $\textbf {a}$ and $\textbf {b}$ make angles $\theta_1$ and $\theta_2$ respectively with the positive direction of $x$-axis and also assume that $\theta_1 > \theta_2$. Then $\theta = \theta_1 – \theta_2$. Also $\tan \theta_1 = \frac {y} {x}$ and $\tan {\theta_2} = \frac {w} {z}$. Then
$$\tan \theta = \frac {\tan \theta_1 – \tan \theta_2} {1 + \tan \theta_1 \tan \theta_2}.$$
Putting the values of $\tan \theta_1$ and $\tan \theta_2$ and then simplifying we get
$$\tan \theta = \frac {yz-xw} {xz+yw}.$$ Since $\theta$ is acute so
$$\cos \theta = \frac {1} {\sqrt {1+{\tan}^2 \theta}} = \frac {xz+yw} {\sqrt {x^2+y^2} \sqrt {z^2+w^2}} = \frac {\langle \textbf {a , b} \rangle} {\|\textbf {a} \| \| \textbf {b} \|}.$$
Best Answer
If the angle $\theta$ between $\bf a,b$ is obtuse, the angle $\pi-\theta$ between $\bf a,-b$ is acute.
$\langle \mathbf a,\mathbf{-b}\rangle=-\langle\mathbf a,\mathbf b\rangle=\|\mathbf a\|\|\mathbf{-b}\|\cos(\pi-\theta)\implies\langle \mathbf a,\mathbf b\rangle=\|\mathbf a\|\|\mathbf b\|\cos(\theta)$.
When $2\pi\ge\theta>\pi,2\pi-\theta\in[0,\pi)$ is the angle between the vectors $\mathbf a,\mathbf b$.
$\langle\mathbf a,\mathbf b\rangle=\|\mathbf a\|\|\mathbf b\|\cos(2\pi-\theta)=\|\mathbf a\|\|\mathbf b\|\cos(\theta)$