May 16th, 2009, 03:37 AM  #1 
Member Joined: May 2009 Posts: 43 Thanks: 0  hard trig!!!
hi there, please help me with this one.. thank's 
May 16th, 2009, 05:24 AM  #2 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: hard trig!!!
Expand sin(2x). Use a common identity to write cos^2(x) in terms of sin(x). Collect like terms and you'll have a cubic equation in sin(x). Look up Cardano's algorithm for solving the cubic [approximately.]

May 16th, 2009, 06:47 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 
All three roots of the cubic are real. For the positive root, one gets 
May 16th, 2009, 07:16 AM  #4 
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re:
Real or not, are they then found readily, or from the general algorithm attributed to Cardano? http://mathworld.wolfram.com/CubicFormula.html My first impression is that there is no simple way to approach the solution other than this, or the application of expensive symbolic algebra programs. Am I incorrect? In any event, over the decades, I've come across many, many postings in newsgroups and forums that eventually boil down to the cubic, and "how do I solve this one?" Sometimes, not always, but often enough, the real agenda is to present the problem of the general cubic one way or another. 
May 16th, 2009, 09:32 AM  #5 
Member Joined: May 2009 Posts: 43 Thanks: 0  Re: hard trig!!!
first of all, thank's for the help. i think there's a simpler solution, because we didn't study that formula (cube) maybe someone can think about other solution? thank's a lot! 
May 16th, 2009, 11:12 AM  #6  
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: hard trig!!! Quote:
That looks like a cubic equation in sin(x) to me. Letting sin(x) = u... No nice round numbers.  
May 16th, 2009, 11:24 AM  #7 
Member Joined: May 2009 Posts: 43 Thanks: 0  Re: hard trig!!!
i'm so sorry... my teacher gave us a bad copy so i didn't notice but you're right: it's not a "2" in the beginning. it's 2(sqrt)3 the rest is o.k. again i'm truly sorry and i would really appreciate your help again.. thank you 
May 16th, 2009, 11:34 AM  #8 
Member Joined: May 2009 Posts: 43 Thanks: 0  Re: hard trig!!!
i'm very sorry.. this is the final and the correct one.. if it's a bother i'll understand if you don't want to help me. 
May 16th, 2009, 03:09 PM  #9  
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: hard trig!!! Quote:
sin(2x) = 2sin(x)cos(x) sin^2(x) + cos^2(x) = 1 I've not worked through it [I'd rather you did], but suspect you still wind up with a cubic.  
May 16th, 2009, 05:05 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 
There was another change in the first term, making the problem much more interesting. Can you finish from there, themanandthe? If so, what solutions do you find? 

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