Proving that $\langle \textbf {a,b} \rangle = \|\textbf {a}\| \|\textbf {b}\| \cos \theta$.

Question

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

$$\cos \theta = \frac {1} {\sqrt {1+{\tan}^2 \theta}} = \frac {xz+yw} {\sqrt {x^2+y^2} \sqrt {z^2+w^2}} = \frac {\langle \textbf {a , b} \rangle} {\|\textbf {a} \| \| \textbf {b} \|}.$$

Answer 1

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 $\mathbf a,\mathbf b$.

$\langle\mathbf a,\mathbf b\rangle=\|\mathbf a\|\|\mathbf b\|\cos(2\pi-\theta)=\|\mathbf a\|\|\mathbf b\|\cos(\theta)$

Answer 2

It follows from the law of cosines. Note that the polarization identity holds: $$ \lVert a+b\rVert^2-\lVert a-b\rVert^2 = 4\langle a,b\rangle. $$ Now, consider the parallelogram determined by 4 vertices $0,a,b,a+b$. We can see that that by the law of cosines, it holds $$ \lVert a+b\rVert^2 = \lVert a\rVert^2+\lVert b\rVert^2 -2\lVert a\rVert\lVert b\rVert\cos(\pi-\theta)=\lVert a\rVert^2+\lVert b\rVert^2 +2\lVert a\rVert\lVert b\rVert\cos\theta, $$and $$ \lVert a-b\rVert^2 = \lVert a\rVert^2+\lVert b\rVert^2 -2\lVert a\rVert\lVert b\rVert\cos\theta, $$ where $\theta$ is the angle between $a$ and $b$. Thus we have $$ \langle a,b\rangle = \lVert a\rVert\lVert b\rVert\cos\theta, $$ as desired.

Answer 3

The proof based on the law of cosines works for obtuse angles as well as acute angles. In the figure below, the law of cosines tells us that \begin{equation} \|y-x\|^2 = \|x\|^2 + \|y\|^2 - 2 \|x\| \|y\| \cos(\theta). \end{equation} Notice that when $\theta = \pi/2$, the law of cosines reduces to the Pythagorean theorem.

We can expand the left-hand side of (1) as follows: \begin{align*} \|y - x \|^2 &= \langle y - x, y - x\rangle \\ &= \langle y - x, y \rangle - \langle y - x, x \rangle \\ &= \langle y,y \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle x, x \rangle \\ &= \| x \|^2 + \|y\|^2 - 2 \langle x, y \rangle. \end{align*} Now comparing the left-hand side of (1) (in expanded form) with the right-hand side, and canceling like terms from both sides, we discover that $\langle x, y \rangle = \|x\| \|y\| \cos(\theta)$.