My Math Forum hard trig!!!

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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 $sin(x)\,=\,(\sqrt{13}/3)\cos(\cos^{\small{-1}}(-(11/26)/\sqrt{13})/3)\,-\,1/6.$
 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:
 Originally Posted by themanandthe i think there's a simpler solution, because we didn't study that formula (cube)
Then you have something wrong in the typing of the question, or were given wrong information. That's the beginning and end of it. There are three solutions within certain bounds. If you graph the relation [y = , not 0 = ...] it extends left and right, then look for intersections with the x=axis. I will take the time to show here why I say it is cubic, but then will not pursue it further than that.

$\begin{array}{l}
2\sin (2x)\cos (x) + 4\sin (x)\cos ^2 (x) - 3\sin ^2 (x) + \cos ^2 (x) = 0 \\
4\sin (x)\cos ^2 (x) + 4\sin (x)\cos ^2 (x) - 3\sin ^2 (x) + (1 - \sin ^2 (x) = 0 \\
8\sin (x)\cos ^2 (x) - 3\sin ^2 (x) + (1 - \sin ^2 (x)) = 0 \\
8\sin (x)(1 - \sin ^2 (x)) - 3\sin ^2 (x) + (1 - \sin ^2 (x)) = 0 \\
8\sin ^3 (x) + 4\sin ^2 (x) - 8\sin (x) - 1 = 0 \\
\end{array}$

That looks like a cubic equation in sin(x) to me. Letting sin(x) = u...

$8u^3 + 4u^2 - 8u - 1= 0$

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:
 Originally Posted by themanandthe 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.
You have already been helped. Except for the slight change, follow the steps in the example given. See what you can do there first, and come back when stuck. If you can't expand sin(2x) or convert cosine to sine where needed, then you need to review those identities first:

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. $2sqrt3\sin(2x)\sin(x)\,+\,4\sin(x)\cos^2(x)\,-\,3\sin^2(x)\,+\,\cos^2(x)\,=\,0$ $2sqrt3\sin(2x)\sin(x)\,+\,2\sin(2x)\cos(x)\,-\,3\sin^2(x)\,+\,\cos^2(x)\,=\,0$ $2\sin(2x)(sqrt3\sin(x)\,+\,\cos(x))\,-\,(\sqrt3\sin(x)\,-\,\cos(x))(\sqrt3\sin(x)\,+\,\cos(x))\,=\,0$ Can you finish from there, themanandthe? If so, what solutions do you find?

 Tags hard, trig

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Singularity Calculus 2 September 30th, 2012 04:38 PM lechatelier Trigonometry 5 July 7th, 2012 12:02 PM panky Calculus 8 August 19th, 2011 02:34 PM jack001 Algebra 4 September 9th, 2009 06:57 AM frosty Calculus 1 August 20th, 2009 08:41 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top