I have a constant X that is the value of an angle. I want to check if the angle formed by 2 given vectors is less than the value X. I know it's possible to calculate the angle using the arccos of the dot product of the vectors divided by the product of the magnitude of the vectors, however I want to do this without applying trigonometric functions to the vectors. Using trigonometric functions on the constant X is fine.
My thoughts were that the dot product of the vectors divided by the product of the magnitude of the vectors gives me the cosine of that angle so if I do the cosine of the angle X maybe I can compare these 2 cosines. Is this viable?
Out of curiosity, is this possible without using trigonometric functions at all?
Solution involving complex numbers
Assume the given vectors $\vec{u},\vec{v}\;$ are represented by the numbers $z_ {u}=a+ib,\;z_{v}=c+id.\;$
Then the oriented angle $\theta$ from $\vec{u}$ to $\vec{v}$ is $$\arg \frac {c+id}{a+ib}.$$ Compute $$\frac {c+id}{a+ib}\cdot \frac{a-ib}{a-ib}=\frac{ac+bd}{a^2+b^2}+i\frac{ad-bc}{a^2+b^2}$$ *Assume $\vec{u}\not \perp\vec{v}.\;$ From the signs of the real and imaginary parts, one can locate $\theta$ into $\left(-\frac \pi2, \frac \pi2\right)\; \text{or}\; \left(\frac \pi2, \frac {3\pi}{2}\right).$
Further, $$\frac{\text{imaginary part}}{\text{real part}}=\frac{ad-bc}{ac+bd}=\tan\theta.$$ To finish, it suffices to locate the given angle $X$ into one of the above intervals and use the monotony of $\;\tan\;$ on this interval.
Note *
Denote $\mathcal{R}=\left\{\frac{(2k-1)\pi}{2}, k =0,1 \right\}.\;$ If $\theta\in \mathcal{R}\; \text{or}\; X \in \mathcal{R},\;$ we have a trivial case easy to solve.