Solving equations in sin and cos theta

  • #1
holulumaster
2
0
Hi,
I have been trying to solve some mechanics problems and I can come up with the right equation, but I don't know how to solve it...
For example:

490.5cos(theta) - 3600cos(theta)sin(theta) + 1800sin(theta) = 0

i need to solve for theta here
its one equation and one unknown, which means it is solvable..
but how?

thanks
 

Answers and Replies

  • #2
nicksauce
Science Advisor
Homework Helper
1,272
5
My suggestion would be to write cos(theta) = x, sin(theta) = sqrt(1-x^2), and simplify. Then you'll have a polynomial equation to solve.
 
  • #3
Gerenuk
1,034
4
That substitution seems to give a 4th order polynomial, which is very impractical.

I rather suggest using
[tex]\cos\theta=\Re (e^{i\theta})[/tex]
and
[tex]\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}[/tex]
which gives you a quadric equation for [itex]e^{i\theta}[/itex]
 
  • #4
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,145
166
That substitution seems to give a 4th order polynomial, which is very impractical.

I rather suggest using
[tex]\cos\theta=\Re (e^{i\theta})[/tex]
and
[tex]\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}[/tex]
which gives you a quadric equation for [itex]e^{i\theta}[/itex]

But that would not give a quadratic polynomial in eiθ, due to the e-iθ and Re[eiθ]terms.

p.s...

Moderator's note, let's let the OP respond with his/her progress on the problem before offering more help.
 
  • #5
Gerenuk
1,034
4
It would give a quadric equation, because that terms cancels.

Yet my method doesn't work, because if an equation is valid for the real part only, you cannot require it to be true for without the Re() operator :(
Because then the angle would be complex.

Anyway. My method doesn't work the way I wanted...

Unless someone sees a way to solve
[tex]\Re(z^2+2az-1)=0[/tex]
with [itex]|z|=1[/itex] and a a complex number.
 
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