My Math Forum Factoriser, spent 5 hours with it
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 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: 1-sin(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: 7,681 Thanks: 2659 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, 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: 20,968 Thanks: 2217 The above method leads to (sin(x) + 1)(2sin²(x) - 2sin(x) + 1). Thanks from SkyCod
December 11th, 2018, 11:42 AM   #5
Newbie

Joined: Dec 2018

Posts: 14
Thanks: 0

Math Focus: mainly algebra
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 01:24 PM.

December 11th, 2018, 12:15 PM   #6
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,016
Thanks: 1600

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 01:24 PM.

 December 11th, 2018, 12:18 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 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: 7,681
Thanks: 2659

Math Focus: Mainly analysis and algebra
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 01:23 PM.

December 11th, 2018, 01:22 PM   #9
Global Moderator

Joined: Dec 2006

Posts: 20,968
Thanks: 2217

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 Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post helen510 Geometry 7 April 3rd, 2018 03:09 AM puppypower123 Calculus 1 March 27th, 2017 03:51 PM Pumaftw Elementary Math 2 October 6th, 2014 09:17 AM sharp Algebra 9 December 21st, 2010 03:39 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top