Solving Sin4x: How to Expand and Simplify Trigonometric Equations

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Homework Help Overview

The discussion revolves around solving the equation sin4θ = cos2θ within the interval -π < θ < π. Participants are exploring methods to expand and simplify the trigonometric expressions involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to expand sin4θ using the angle addition formula but finds the process cumbersome and questions its correctness. Other participants suggest using known identities like sin2φ = 2sinφcosφ and De Moivre's formula. There are discussions about when it is valid to cancel terms in the equation, particularly concerning the cos2θ term.

Discussion Status

Participants are actively sharing insights and methods, with some guidance provided on the use of trigonometric identities. There is a mix of interpretations regarding the steps to take next, and while some participants express confusion about the results compared to textbook answers, no consensus has been reached on a definitive approach.

Contextual Notes

There are indications of differing results from textbook solutions, leading to questions about the validity of certain steps taken in the problem-solving process. Participants are also considering the implications of canceling terms in the context of the equation.

breen155
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Hey, i need help with a problem I'm having I'm not sure about an expansion

Homework Statement



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

2. The attempt at a solution
I have tried putting sin4\theta as sin(2\theta+2\theta) and expanding to get sin2\thetacos2\theta + cos2\thetasin2\theta. 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
 
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Ever heard of: \sin 2\phi = 2\sin \phi \cos \phi ?
 
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
 
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 :confused:
 
You don't need to keep on expanding. Substituting once gives you a \cos2\theta term on both sides. You can cross those away (under certain conditions).
 
so is it 2sin2x = 1 ? :)
 
Yes, correct, but only when you are allowed to cross out the \cos 2\theta term. Any idea when that move is not allowed..?
 
not too sure is it when they are being added or subtracted?
 
Be careful with your next move... ;)

http://pix.motivatedphotos.com/2008/11/8/633617527281144116-dividebyzero.jpg
 
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  • #10
:P in the end i get sinxcosx =0.25 but in not sure how to get the answer out of it :P
 
  • #11
Stop expanding. The previous equation was as far as you needed to go. You have to know this one by heart: \sin 2\theta = \frac{1}{2}.

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

If not, look it up ;)
 
  • #12
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
 
  • #13
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.
 
  • #14
I finally got it, thanks for the help everyone :)
 

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