I'm confused about why does the dot product between two unit vectors equal the cosine on the angle between them
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Suppose the two unit vectors you have are given by $\langle \cos\theta,\sin\theta \rangle$ and $\langle \cos\phi,\sin\phi \rangle$. These are unit vectors because $\cos^2x+\sin^2x=1$ for any $x$. By definition, the angle between these two vectors would be $\theta-\phi$. Now, write the dot product: $$ \begin{split} \langle\cos\theta,\sin\theta\rangle\cdot\langle\cos\phi,\sin\phi\rangle &= \cos\theta\cos\phi+\sin\theta\sin\phi\\ &=^* \cos(\theta-\phi)\end{split}$$ Where $=^*$ follows because of the angle subtraction identity for cosine. Thus the dot product of two unit vectors is the cosine of the angle between them.
YiFan Tey
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Consider the $\mathbb{R}^2$ first. I suppose you know what a cosine of two vector is there.
Use goniometry and some vector algebra etc. in the plane to see that there the inner product (aka dot product) indeed equals the cosine.
Then mathematicians decided that this allowed for a nice definition for the cosine of two vectors in any vector space with a dot product. So this definition generalises the known fact for the plane to a definitonal truth in all dimensions. We want orthogonal = dot product $0$, which fits the cosine idea too. If the vectors are equal up to sign, we get the expected $-1 = \cos 180^\circ$ etc.
Henno Brandsma
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I have read something about the unitary circle,where the unit vectors are (cos Ø,sin Ø) and then multiplied each other. Isn't that correct? – Ivan C. Oct 24 '18 at 22:23
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@IvanC. On the unit circle, the points have coordinates $(\cos \theta, \sin \theta)$ for $\theta \in [0,2\pi)$. You could look at angles between two of them and apply some gonio rules. – Henno Brandsma Oct 24 '18 at 22:26