Solving Sin4x: How to Expand and Simplify Trigonometric Equations

[breen155]

Hey, i need help with a problem I'm having I'm not sure about an expansion

Homework Statement

Solve the following equation for -[tex]\pi[/tex]<[tex]\theta[/tex]<[tex]\pi[/tex]
sin4[tex]\theta[/tex] = cos2[tex]\theta[/tex]
I'm not sure about how to expand sin4[tex]\theta[/tex]

2. The attempt at a solution
I have tried putting sin4[tex]\theta[/tex] as sin(2[tex]\theta[/tex]+2[tex]\theta[/tex]) and expanding to get sin2[tex]\theta[/tex]cos2[tex]\theta[/tex] + cos2[tex]\theta[/tex]sin2[tex]\theta[/tex]. However it just becomes very messy, I was just wondering if i am doing this correctly as the answer i receive is different to the one in my textbook

 

[xepma]

Ever heard of: [tex]\sin 2\phi = 2\sin \phi \cos \phi[/tex] ?

 

[peterjaybee]

Hia,

I remember having to do this same problem in my first year at uni, but i can't remember the exact method. You have to use something called De Moivre's formula (http://en.wikipedia.org/wiki/De_Moivre's_formula).

Hope that helps,

Peter

 

[breen155]

yeah, i get
sin4x = 2sin2xcos2x and from then on i do
2[(2sinxcosx)(cos²x-sin²x)]
i then carry on expanding and end up with 4sinxcos³x-4sin³xcosx
im not sure how to carry on from there

 

[xepma]

You don't need to keep on expanding. Substituting once gives you a [tex]\cos2\theta[/tex] term on both sides. You can cross those away (under certain conditions).

 

[breen155]

so is it 2sin2x = 1 ? :)

 

[xepma]

Yes, correct, but only when you are allowed to cross out the [tex]\cos 2\theta[/tex] term. Any idea when that move is not allowed..?

 

[breen155]

not too sure is it when they are being added or subtracted?

 

[xepma]

Be careful with your next move... ;)

http://pix.motivatedphotos.com/2008/11/8/633617527281144116-dividebyzero.jpg

 

[breen155]

:P in the end i get sinxcosx =0.25 but in not sure how to get the answer out of it :P

 

[xepma]

Stop expanding. The previous equation was as far as you needed to go. You have to know this one by heart: [tex]\sin 2\theta = \frac{1}{2}[/tex].

Put differently, for what value of [tex]\phi[/tex] do we have [tex]\sin\phi=\frac{1}{2}[/tex]?

If not, look it up ;)

 

[breen155]

if i place that in my formula it doesn't give me the same answer as my textbook though
if sinx = 0.5 then 0.5cosx=1/4 then it gets cosx = .5
this is pie/3 however the answers in my textbook say pie/12 pie/4 etc

 

[rock.freak667]

for sin2x=0.5 =>2x=pi/6 and the other angles in the range ( find what x equals to)

for cos2x=0 => 2x=pi/2,other angles in the range you want (find x)

combine these answers.

 

[breen155]

I finally got it, thanks for the help everyone :)