Solving for Theta, wolfram doesn't give anything useful

  • Thread starter Thread starter physstudent.4
  • Start date Start date
  • Tags Tags
    Theta
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
physstudent.4
Messages
12
Reaction score
0

Homework Statement


I need to write a function of theta in terms of a particular variable. I just can't seem to figure it out; the only solution I can come up with is when the aforementioned variable is equal to 0 or 1. I'm using Q to denote theta, and b is the variable.

sin^3(Q)/cos(Q)=b^2


Homework Equations


Any and all trig identities.
I've used sin^2(x)+cos^(x)=1, sin(2x)=2sin(x)cos(x), I haven't stumbled into any work that led me to use other identities!

The Attempt at a Solution


Around 5 pages of scratch work, all for naught.
Any relevant trig identities I might be missing would be nice to know, if anyone figures this out soon! Thanks!
 
Physics news on Phys.org
physstudent.4 said:

Homework Statement


I need to write a function of theta in terms of a particular variable. I just can't seem to figure it out; the only solution I can come up with is when the aforementioned variable is equal to 0 or 1. I'm using Q to denote theta, and b is the variable.

sin^3(Q)/cos(Q)=b^2

Homework Equations


Any and all trig identities.
I've used sin^2(x)+cos^(x)=1, sin(2x)=2sin(x)cos(x), I haven't stumbled into any work that led me to use other identities!

The Attempt at a Solution


Around 5 pages of scratch work, all for naught.
Any relevant trig identities I might be missing would be nice to know, if anyone figures this out soon! Thanks!

sin3(θ)/cos(θ) = 1/(cot(θ)csc2(θ)) = 1/(cot(θ)+cot3(θ))

but I don't think this will help all that much.
 
You can reduce it to a polynomial equation. Square both sides. Replace cos(Q)^2 by 1-sin(Q)^2. Now let P=sin(Q)^2. You'll get a cubic equation in P. You CAN solve a cubic exactly, but the solutions are so complicated as to be almost useless. You will have a little advantage here because there is no P^2 term. But it's still pretty complicated. Stuff like this is usually handled numerically, not analytically.