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

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 $1$.

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  • $\begingroup$ As already noticed we can use of course vector interpretation to obtain the same results we can obtain by trigonometric identitie (which indeed can be also proved in that way) but in that case we can't obtain the full solution by vector interpretation. $\endgroup$ – gimusi Sep 5 '18 at 23:51
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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 $a+a = 2a$.

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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.
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