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#1 2007-07-23 09:44:41

Bigbro
Guest

Trig Equation Help!

cos^2 4x- sin^2 4x =0

Help plz

#2 2007-07-23 09:56:12

Bigbro
Guest

Re: Trig Equation Help!

Clarification

cos²4x-sin²4x=0

#3 2007-07-23 11:17:52

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Trig Equation Help!

well the easiest way to look at this is:

let u = 4x

cos^2(u) = sin^2(u)

this says that cos(u) = ± sin(u). The reason for this is because even if they're signs are different, they are equal when you square them so we must consider both.

It turns out that if you look at the angles 45, 45 + 90, 45 + 180 and 45 + 270, the signs and cosines of each particular angle are either equal or opposites, which fits the equation cos(u) = ± sin(u).  We could consider more values, namely 45 + 360, 45 + 450 but we're just looping around again so these solutions are redundant.

Since we're using radian form, the four answers become

u = pi/4, pi/4 + pi/2, pi/4 + pi, and pi/4 + 3pi/2

so u = pi/4, 3pi/4, 5pi/4 and 7pi/4

since u = 4x we have

4x = pi/4
4x = 3pi/4
4x = 5pi/4
4x = 7pi/4

so x = pi/16, 3pi/16, 5pi/16, or 7pi/16. All of these are solutions to the equation. This is why i hate trig equations. sad


A logarithm is just a misspelled algorithm.

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#4 2007-07-23 12:20:10

Bigbro
Guest

Re: Trig Equation Help!

Thanks alot smile

#5 2007-07-23 13:33:06

bossk171
Member
Registered: 2007-07-16
Posts: 305

Re: Trig Equation Help!

mikau wrote:

so x = pi/16, 3pi/16, 5pi/16, or 7pi/16. All of these are solutions to the equation. This is why i hate trig equations. sad

My trig teatcher always demanded we write

in which k is a whole number. I always thought it was over kill, but he demanded "absolute truth."


There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

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#6 2007-07-23 18:19:24

reddog
Member
Registered: 2007-07-23
Posts: 7

Re: Trig Equation Help!

and
can't be 0 in the same time. So we can divide the equation by
and we get

Then

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#7 2007-07-23 19:01:07

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Trig Equation Help!

yeah, you can always throw in some unspecified integer. The benefit there is you litterally cover all possible values of x.

Its probably safer to do that come to think of it. some teachers are picky.


A logarithm is just a misspelled algorithm.

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#8 2007-07-23 20:22:58

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Trig Equation Help!

Possibly redundant now, but from here: cos²4x-sin²4x=0

I'd recognise that as a Trig. identity and turn it into cos 8x=0, which then solves in the same way as the others.


Why did the vector cross the road?
It wanted to be normal.

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#9 2007-07-24 00:28:08

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Trig Equation Help!

I'd recognise that as a Trig. identity and turn it into cos 8x=0, which then solves in the same way as the others.

Are you recognizing it as a trig identity for the same reason I am now?  Cause George posted it in that other topic?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2007-07-24 00:32:25

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Trig Equation Help!

Not really, I just knew that one. My most recent module relied quite heavily on them, so I can remember all the basic ones.


Why did the vector cross the road?
It wanted to be normal.

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