My Math Forum Confusing Pre-Calculus Problem. Can Anyone Counter This?

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 19th, 2013, 04:03 PM #1 Newbie   Joined: Jan 2013 Posts: 3 Thanks: 0 Confusing Pre-Calculus Problem. Can Anyone Counter This? Question: Sin 4x = -2 Sin 2x What I did: = 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula = 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula =2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2 = (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx =2-2sin^2 x = 0 Added 1 on both sides. =2(1-sinx)(1+sinx) = 0 Factored 1- sinx = 0 & sinx + 1 =0 Obvious method sinx = 1 & sinx = -1 " x= pi/2 U x = 3pi/2 Unit circle. But however, the answer key says this is wrong. Others' solutions: Sin 4x = -2 sin 2x = 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula = 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x 2sin 2x = 0 & cos2x = -1 sin 2x = 0 & cos 2x = -1 0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2 x= 0, pi, pi/2, 3pi/2 Does anyone know what I did wrong? ( the first solution)
January 19th, 2013, 04:23 PM   #2
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: Confusing Pre-Calculus Problem. Can Anyone Counter This?

Quote:
 Originally Posted by bboyinmartin Question: Sin 4x = -2 Sin 2x What I did: = 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula
Yes, that is correct. Although, personally, I wouldn't have use the "double angle formula" on the right. I would have stopped with a formula in "2x" on both sides.

Quote:
 = 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula cos(2x)= 1+ 2 cos^2(x)) =2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2 = (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx
IF $2 sin(x)cos(x)\ne 0$. Otherwise you cannot do that. You need to check those cases, sin(x)= 0 or cos(x) 0, separately.

Quote:
 [tex]=2-2sin^2 x = 0 Added 1 on both sides. =2(1-sinx)(1+sinx) = 0 Factored 1- sinx = 0 & sinx + 1 =0 Obvious method sinx = 1 & sinx = -1 " x= pi/2 U x = 3pi/2 Unit circle. But however, the answer key says this is wrong. Others' solutions: Sin 4x = -2 sin 2x = 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula = 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x 2sin 2x = 0 & cos2x = -1 sin 2x = 0 & cos 2x = -1 0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2 x= 0, pi, pi/2, 3pi/2
All cases in which either sin(x)= 0 or cos(x)= 0, and therefore sin(x)cos(x)= 0.

Quote:
 Does anyone know what I did wrong? ( the first solution)
You can't divide by 0!

January 19th, 2013, 06:33 PM   #3
Newbie

Joined: Jan 2013

Posts: 3
Thanks: 0

Re: Confusing Pre-Calculus Problem. Can Anyone Counter This?

Quote:
Originally Posted by HallsofIvy
Quote:
 Originally Posted by bboyinmartin Question: Sin 4x = -2 Sin 2x What I did: = 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula
Yes, that is correct. Although, personally, I wouldn't have use the "double angle formula" on the right. I would have stopped with a formula in "2x" on both sides.

Quote:
 = 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula cos(2x)= 1+ 2 cos^2(x)) =2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2 = (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx
IF $2 sin(x)cos(x)\ne 0$. Otherwise you cannot do that. You need to check those cases, sin(x)= 0 or cos(x) 0, separately.

[quote:2rd9xf30][tex]=2-2sin^2 x = 0 Added 1 on both sides.
=2(1-sinx)(1+sinx) = 0 Factored
1- sinx = 0 & sinx + 1 =0 Obvious method
sinx = 1 & sinx = -1 "
x= pi/2 U x = 3pi/2 Unit circle.

But however, the answer key says this is wrong.
Others' solutions:
Sin 4x = -2 sin 2x
= 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula
= 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x
2sin 2x = 0 & cos2x = -1
sin 2x = 0 & cos 2x = -1
0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2

x= 0, pi, pi/2, 3pi/2
All cases in which either sin(x)= 0 or cos(x)= 0, and therefore sin(x)cos(x)= 0.

Quote:
 Does anyone know what I did wrong? ( the first solution)
You can't divide by 0![/quote:2rd9xf30]

Oh, okay. Thanks!

 January 19th, 2013, 06:48 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,206 Thanks: 513 Math Focus: Calculus/ODEs Re: Confusing Pre-Calculus Problem. Can Anyone Counter This? I would use the strategy suggested by [color=#0000FF]HallsofIvy[/color]: $\sin(4x)=-2\sin(2x)$ Use the double-angle identity for sine on the left, and arrange as: $2\sin(2x)\cos(2x)+2\sin(2x)=0$ Factor: $2\sin(2x)\(\cos(2x)+1)=0$ We have two cases to consider (where $k\in\mathbb{Z}$): i) $\sin(2x)=0\,\therefore\,2x=k\pi\,\therefore\,x=\fr ac{\pi}{2}k$ ii) $\cos(2x)=-1\,\therefore\,2x=(2k+1)\pi\,\therefore\,x=\frac{\ pi}{2}(2k+1)$ However, we see that all of the solutions from the second case are part of the first, so we may simply state: $x=\frac{\pi}{2}k$

 Tags confusing, counter, precalculus, problem

,

,

,

### mathematics 2sinx formula

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post YoungMoustache Elementary Math 1 September 20th, 2012 07:01 PM cherryperry Elementary Math 1 May 29th, 2010 02:29 PM suomik1988 Algebra 4 October 28th, 2009 02:27 AM beatboxbo Calculus 5 October 25th, 2007 01:43 AM VDSL_2 Algebra 3 September 11th, 2007 10:56 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top