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 December 11th, 2018, 06:57 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: 1-sin(x).cos(2x) Thanks. Last edited by skipjack; December 11th, 2018 at 08:59 AM. December 11th, 2018, 07:15 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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. Thanks from SkyCod December 11th, 2018, 08:17 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 08:58 AM. December 11th, 2018, 08:52 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 The above method leads to (sin(x) + 1)(2sin²(x) - 2sin(x) + 1). Thanks from SkyCod December 11th, 2018, 12:42 PM   #5
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Quote:
 Originally Posted by skipjack The above method leads to (sin(x) + 1)(2sin²(x) - 2sin(x) + 1).
How come it does? And even if it did, it's still not a complete factorisation.

Last edited by skipjack; December 11th, 2018 at 02:24 PM. December 11th, 2018, 01:15 PM   #6
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Quote:
 Originally Posted by SkyCod How come it does? And even if it did, it's still not a complete factorisation.
$1-\sin{x}\cos(2x)$

$1-\sin{x}(1-2\sin^2{x})$

$1 - \sin{x} + 2\sin^3{x}$

$(1 + \sin^3{x}) - (\sin{x} - \sin^3{x})$

$(1+\sin{x})(1-\sin{x}+\sin^2{x}) - \sin{x}(1-\sin^2{x})$

$\color{red}{(1+\sin{x})}(1-\sin{x}+\sin^2{x}) - \sin{x}(1-\sin{x})\color{red}{(1+\sin{x})}$

$\color{red}{(1+\sin{x})}\bigg[(1-\sin{x}+\sin^2{x}) - \sin{x}(1-\sin{x})\bigg]$

$(1+\sin{x})\bigg[1-\sin{x}+\sin^2{x} - \sin{x}+\sin^2{x}\bigg]$

$(1+\sin{x})(1-2\sin{x}+2\sin^2{x})$

that's it ...

Last edited by skipjack; December 11th, 2018 at 02:24 PM. December 11th, 2018, 01:18 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 You asked for help in factorizing, not for a complete factorization, but skeeter's provided more detail for you. December 11th, 2018, 01:45 PM   #8
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Quote:
 Originally Posted by SkyCod How come it does? And even if it did, it's still not a complete factorisation.
It is a complete factorisation with real coefficients. The polynomial $2x^2-2x+1$ is irreducible because $b^2 - 4ac = 4 - 8 = -4 <0$

Last edited by skipjack; December 11th, 2018 at 02:23 PM. December 11th, 2018, 02:22 PM   #9
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Quote:
 Originally Posted by SkyCod How come it does?
\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*} Tags factoriser, hours, spent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post helen510 Geometry 7 April 3rd, 2018 04:09 AM puppypower123 Calculus 1 March 27th, 2017 04:51 PM Pumaftw Elementary Math 2 October 6th, 2014 10:17 AM sharp Algebra 9 December 21st, 2010 04:39 PM

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