0

Given $a*\sin(\theta) + b*\cos(\theta) < l$, how can I find valid values for $\theta$ in the range of $0..2\pi$?

I was able to solve a similar equation using Solving $a \sin\theta + b \cos\theta = c$, but for my actual problem I am not really sure what to do.

Aart Stuurman
  • 175
  • 2
    Where do $a$ and $b$ live? First use your previous work at solving $a \sin \theta + b\cos \theta = c$, but replace c everywhere by l. That will establish the boundary of yourinequality. – amWhy Mar 16 '20 at 13:30
  • 1
    You can write $\cos\theta$ as $\sqrt{1-\sin^2 \theta}$ to get a quadratic inequality in $\sin\theta$. It should be bounded between 2 values. – Vishu Mar 16 '20 at 14:08

1 Answers1

0

First, you can say $n=\sqrt{a^2+b^2}$ and then multiply and divide both sides, obtaining: $$\frac{a}{n}\sin(\theta)+\frac{b}{n}\cos(\theta)<l\leftrightarrow\frac{a}{\sqrt{a^2+b^2}}\sin(\theta)+\frac{b}{\sqrt{a^2+b^2}}\cos(\theta)<l$$ From here, you can write: $$\sin\left(\theta+\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)\right)<l$$ Solving this is very simple, in fact: $$\arcsin(l)-\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)+2k\pi<\theta<\arcsin(l)-\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)+2k\pi+\pi$$

Matteo
  • 6,581