trig identity issue

glenn101 · 2009-09-12 14:23:21

Hi all just one hopefully quick question which is really bothering me

Solve sin2x=sinx over [0,2pi]

sin2x=2sinxcosx from double angle formula

2sinxcosx=sinx
divide both sides by 2sinx
cosx=sinx/2sinx
cosx=1/2
x= pi/3 or 5pi/3

this is wrong however, what is it that is incorrect?

bobbym · 2009-09-12 16:37:00

Hi glenn101

The pi /3 and 5 / 3 pi are fine you just missed 0, and 2 pi, I believe the division by 2 sin(x) is the culprit.

Look at this simple example:

x^2 - 2x =0

If you unwisely solve this by dividing by x you get

x - 2 = 0

x = 2

Notice that you lost the x = 0 solution.

Last edited by bobbym (2009-09-12 16:58:14)

glenn101 · 2009-09-13 19:12:17

Thanks for the reply bobbym:)
thats interesting your example with x^2-2x=0
because technically you ca divide both sides by zero which will make x=2.
So why is it that this results in missing solutions though? You've got me interested with this now.

bobbym · 2009-09-13 19:38:25

Hi glenn101;

Division by zero is a no no. When it is masked because you are dividing by variables it can produce very strange results. Wiping out roots is just one of them, just like the small example I gave. Better than the division would have been to factor x^2 - 2x = x(x-2) = 0 which gets all the roots. The example is just a toy problem but I think that is what is happening in your trig problem. Try to factor it rather than dividing by 2 sin(x).

I believe that it is similar to the process of deflation. One of the roots of your trig equation is 0 you are deflating it by dividing by 2 sin(x) (which also has a root of zero).

Math Is Fun Forum

#1 2009-09-12 14:23:21

trig identity issue

#2 2009-09-12 16:37:00

Re: trig identity issue

#3 2009-09-13 19:12:17

Re: trig identity issue

#4 2009-09-13 19:38:25

Re: trig identity issue

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