How can I solve this system of trigonometric equations analytically? It is from physics class. $$ \begin{cases} 30t\cos{\alpha}=50\\ -30t\sin{\alpha}-4.9t^2=0 \end{cases} $$

  • $\begingroup$ Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms? $\endgroup$ – Josh B. Nov 28 '18 at 17:16

Hint: Squaring both the equations, you will get $900t^2\cos^2{\alpha}=2500\\ 900t^2\sin^2{\alpha}={4.9}^2t^4$.

Note that $\sin^2{\alpha}+\cos^2{\alpha}=1$.

So add both the equations and solve for $t$ using the substitution $t^2=u$.

  • 1
    $\begingroup$ Turn it into a quadratic is the gist of my answer but this is much nicer+faster. $\endgroup$ – Mason Nov 28 '18 at 17:22
  • $\begingroup$ @Mason Thank you. $\endgroup$ – Thomas Shelby Nov 28 '18 at 17:24

$30t\cos{\alpha}=50 \implies t=\frac{5}{3} \sec\alpha$

You can plug this information into the other equation and solve:

$$-30t\sin{\alpha}-4.9t^2=0\implies -30(\frac{5}{3} \sec\alpha)\sin{\alpha}-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} \sec^2\alpha=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} (\tan^2\alpha+1)=0$$

Taking $y=\tan\alpha$ you can solve a quadratic equation.
$$-50(y)-4.9\frac{25}{9} (y^2+1)=0$$

I think you're probably in good shape from here?

  • $\begingroup$ This results in negative alpha but its a real physics problem, so it must be positive. $\endgroup$ – Stepii Nov 28 '18 at 17:27
  • $\begingroup$ The angle cannot be negative? Add $2\pi$? $\endgroup$ – Mason Nov 28 '18 at 17:29
  • $\begingroup$ Adding 2π gives angles greater than 90° $\endgroup$ – Stepii Nov 28 '18 at 17:38
  • $\begingroup$ @Stepii Which also doesn't make sense in the context of the problem? $\endgroup$ – Mason Nov 28 '18 at 17:43
  • $\begingroup$ What did you get for $y$. Maybe I made an error... Here's y $\endgroup$ – Mason Nov 28 '18 at 17:44

Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.

enter image description here

This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $a<0$ then $t>0$

  • $\begingroup$ The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems. $\endgroup$ – Mason Nov 28 '18 at 18:23
  • $\begingroup$ But that solutions (with the angle and time both positive) satisfy the given system $\endgroup$ – Stepii Nov 28 '18 at 18:29

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