Vector equation: $\left<\cos(a), \sin(a)\right> \cdot \left<\cos(a),-\sin(a) \right> = -\cos(a)$

Question

$$\left<\cos(a), \sin(a)\right> \cdot \left<\cos(a),-\sin(a) \right> = -\cos(a)$$

Writing $\textbf{u}=\left<\cos(a), \sin(a)\right>$ and $\textbf{v}=\left<\cos(a), -\sin(a)\right>$. We observe that both u and v are unit vectors, and the angle between them is $2a$. Therefore from

$$\textbf{u}\cdot \textbf{v} = |\textbf{u}||\textbf{v}|\cos(\theta)$$

Where $\theta$ is the angle between the two vectors) we find that the equation is equivalent to

$$\cos(2a) = -\cos(a)$$

Could you explain what he has done so far? I did not get it properly. As a first thing made me confused, how did he obtain the angle between them? It is obvious that $\textbf{u}=\left<\cos(a), \sin(a)\right>$ and $\textbf{v}=\left<\cos(a), -\sin(a)\right>$ are unit vectors, therefore they equal $1$.

Answer 1

It is obvious that $\textbf{u}=\left<\cos(a), \sin(a)\right>$ and $\textbf{v}=\left<\cos(a), -\sin(a)\right>$ are unit vectors, therefore they equal $1$.

Be careful with pronouns. “They” refers to vectors, while $1$ is a scalar. Vectors can't be equal to a scalar. I think what you mean is that $\mathbf{u}$ and $\mathbf{v}$ have length $1$. Keep your language neat to organize your thinking.

But to your main question: If you draw $\mathbf{u}$ and $\mathbf{v}$ in the plane, you see $\mathbf{u}$ is the unit vector in the first quadrant making an angle of $a$ with the positive $x$-axis, and $\mathbf{v}$ is the unit vector in the fourth quadrant making an angle of $a$ with the positive $x$-axis. So the angle between $\mathbf{u}$ and $\mathbf{v}$ is $a+a = 2a$.

Answer 2

There are two ways to represent this:

• Either consider two vectors $u$ and $v$ with coordinates $(\cos(x),\sin(x))$ and $(\cos(x),-\sin(x))$. The angle between them will be $2x$ as the angle between the axis and $u$ is $x$ and the angle between the axis and $v$ is x. They are unit-vectors, so the formula described holds
• Mathematically speaking, $(a,b)\cdot (c,d) = ac + bd$ here, you get a scalar product equal to $\cos^2(x) - \sin^2(x) = \cos(2x)$ so in both ways, you find the same result.