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December 11th, 2018, 05:57 AM  #1 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  Factoriser, spent 5 hours with it
i understand this might be easy even if I spent a lot of time to get a clue, but there is a beginning to everything so here it is if anyone can help me: Factorise: 1sin(x).cos(2x) Thanks. Last edited by skipjack; December 11th, 2018 at 07:59 AM. 
December 11th, 2018, 06:15 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra 
$$\cos 2A = \cos^2 A  \sin^2 A = 1  2\sin^2 A$$ This leads to a cubic in $\sin x$ that can be factorised. 
December 11th, 2018, 07:17 AM  #3 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra 
Yes, I did follow that and with no actual lead if you put it on paper, you can try it.
Last edited by skipjack; December 11th, 2018 at 07:58 AM. 
December 11th, 2018, 07:52 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
The above method leads to (sin(x) + 1)(2sin²(x)  2sin(x) + 1).

December 11th, 2018, 11:42 AM  #5 
Newbie Joined: Dec 2018 From: Canada Posts: 14 Thanks: 0 Math Focus: mainly algebra  How come it does? And even if it did, it's still not a complete factorisation.
Last edited by skipjack; December 11th, 2018 at 01:24 PM. 
December 11th, 2018, 12:15 PM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  Quote:
$1\sin{x}(12\sin^2{x})$ $1  \sin{x} + 2\sin^3{x}$ $(1 + \sin^3{x})  (\sin{x}  \sin^3{x})$ $(1+\sin{x})(1\sin{x}+\sin^2{x})  \sin{x}(1\sin^2{x})$ $\color{red}{(1+\sin{x})}(1\sin{x}+\sin^2{x})  \sin{x}(1\sin{x})\color{red}{(1+\sin{x})}$ $\color{red}{(1+\sin{x})}\bigg[(1\sin{x}+\sin^2{x})  \sin{x}(1\sin{x})\bigg]$ $(1+\sin{x})\bigg[1\sin{x}+\sin^2{x}  \sin{x}+\sin^2{x}\bigg]$ $(1+\sin{x})(12\sin{x}+2\sin^2{x})$ that's it ... Last edited by skipjack; December 11th, 2018 at 01:24 PM.  
December 11th, 2018, 12:18 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
You asked for help in factorizing, not for a complete factorization, but skeeter's provided more detail for you.

December 11th, 2018, 12:45 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra  It is a complete factorisation with real coefficients. The polynomial $2x^22x+1$ is irreducible because $b^2  4ac = 4  8 = 4 <0$
Last edited by skipjack; December 11th, 2018 at 01:23 PM. 
December 11th, 2018, 01:22 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  $\displaystyle \begin{align*}2\sin^3(x)  \sin(x) + 1 &= 2\sin(x)(\sin^2(x)  1) + \sin(x)+ 1 \\ &= 2\sin(x)(\sin(x) + 1)(\sin(x) 1) + \sin(x) + 1 \\ &= (\sin(x) + 1)(2\sin(x)(\sin(x)  1) + 1) \\ &= (\sin(x) + 1)(2\sin^2(x)  2\sin(x) + 1) \\ &= (\sin(x) + 1)((1+ i)\sin(x)  1)((1  i)\sin(x)  1)\end{align*}$ 

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