Printable View
Hi.
Can anyone help me find the angles (values of t) where:
25*sin(t) = 22*sin(2t) - 18*cos(2t)
I know the general way to do it is to get all one trig function and form a quadratic then solve it, but I'm not too hot on the identities and I can't seem to get rid of the double angle bits and make everything sine or anything like that!
Help!
Thanks!
25*sin(t) = 22*sin(2t) - 18*cos(2t)Quote:
Originally Posted by TheOnlyNameLeft
sin(2t) = 2*sin(t)*cos(t)
cos(2t) = 1 - 2*sin^2(t)
So
25*sin(t) = 44*sin(t)*cos(t) - 18 + 36*sin^2(t)
Now, the problem is that we need to get rid of the cos(t) term somehow. There is only one reasonable way to do this: we know that sin^2(t) + cos^2(t) = 1, so...
cos(t) = (+/-)sqrt{1 - sin^2(t)}
Thus:
25*sin(t) = (+/-)44*sin(t)*sqrt{1 - sin^2(t)} - 18 + 36*sin^2(t)
Do yourself a favor at this point, let x = sin(t). Then:
25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2
Isolate the square root term:
(+/-)44x*sqrt{1 - x^2} = -36x^2 + 25x + 18
Square both sides:
1936x^2*(1 - x^2) = (-36x^2 + 25x + 18)^2
1936x^2 - 1936x^4 = 1296x^4 - 1800x^3 - 671x^2 + 900x + 324
3232x^4 -1800x^3 - 2607x^2 + 900x + 324 = 0
There's nothing easy about solving this, AND we need to check each x back in the original problem:
25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2
Once that is done, the final solution is
t = asn(x)
I have attached a graph of the polynomial function. I'll let you do the approximations and get the solutions for t.
-Dan
Thanks a lot. :)
Had some attempts but didn't think of getting rid of the cos(t) using the sin^2 + cos^2 = 1 identity!
That's actually a very small part of a dynamics exam question about finding required flywheel sizes. I think I'll have to go visit the lecturer and see if he's expecting us to do things like that in the exam or use the graphs given to estimate the angles where the functions cross!
Thanks again.