In Simmmon's Calculus with Analytic Geometry 3rd edition, on page 299, the following proof for the law of cosines is provided:
$$c^2 = a^2 + b^2 - 2ab \cos \theta$$ The proof is routine if we place the triangle in the $xy$-plane as shown in the figure and apply the distance formula to the vertices $(a \cos \theta, a \sin \theta)$ and $(b, 0)$
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My doubt is: why are the coordinates of the left vertex $(a \cos \theta, a \sin \theta)$? Thanks in advance.
On
Because that vertex is what we get applying a rotation of angle $\theta$ to the point $(a,0)$. And applying such a rotation to a point $(x,y)$ gives us $\bigl(x\cos(\theta)-y\sin(\theta),x\sin(\theta)+y\cos(\theta)\bigr)$.
On
This is just the usual notation of parametric coordinates of a point which at a times proves to be very useful. For more information you can refer to the below link
https://en.m.wikipedia.org/wiki/Parametric_equation
Hope it helps.
You can draw a right-angled triangle to explain this. For a point in the second quadrant, as you have drawn it here, drop a perpendicular from your point $P$ to the $x$-axis, meeting it at $Q$. Then $OPQ$ is right-angled with $\angle POQ=\pi-\theta$. Therefore $|PQ|=a\sin(\pi-\theta)$ and $|OQ|=a\cos(\pi-\theta)$. Then $P$ has coordinates $$(-a\cos(\pi-\theta),a\sin(\pi-\theta))=(a\cos\theta,a\sin\theta).$$ Similar arguments work in each of the other three quadrants.