0
$\begingroup$

How to solve this system of non linear trigonometric equations:

$$\begin{align} A\sin\theta_1+\phantom{5\omega}B\sin\theta_2 &=P \tag{1}\\ 2A\sin\theta_1+\phantom{\omega}5B\sin\theta_2 &=Q \tag{2}\\ A\omega\cos\theta_1+\phantom{5}B\omega\cos\theta_2 &=0 \tag{3}\\ 2A\omega\cos\theta_1+5B\omega\cos\theta_2 &=0 \tag{4} \end{align}$$

$A$, $B$, $\theta_1$, $\theta_2$ are variables, and $\omega$ is a constant.

Can you at least give me a hint on how to proceed?

$\endgroup$
  • 2
    $\begingroup$ Note that $\omega$ in equations $3$ and $4$ is of no use. That is, if it is nonzero. $\endgroup$ – AryanSonwatikar Dec 24 '19 at 3:32
0
$\begingroup$

Following assumptions are to be made: $A, B,$ and $\omega \ne 0$
$$\left(2\right)-2\times\left(1\right)\space\space\space\space 3B\sin\theta_2=Q-2P$$ $$\left(4\right)-2\times\left(3\right)\space\space\space\space\space 3B\cos\theta_2=0\space\space\space\space\space\space\space\space\space\space\space$$ $$5\times\left(1\right)-\left(2\right)\space\space\space\space 3A\sin\theta_1=5P-Q$$ $$5\times\left(3\right)-\left(4\right)\space\space\space\space\space 3A\cos\theta_1=0\space\space\space\space\space\space\space\space\space\space\space$$ This is the hint you requested. Now please proceed.

$\endgroup$
  • $\begingroup$ Thank you very much. $\endgroup$ – Amogh M H Jan 4 at 2:21
0
$\begingroup$

Hint:

Use $(3)$ to simplify $(4)$ and $(1)$ to simplify $(2)$. Square and add to eliminate $\theta_2$. Can you proceed?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.