Jump to content

non intuitive way of solving cos(X)=cos(x+pi/2)


Recommended Posts

as stated' date=' is there a none intuitive way of solving

 

[math']

 

cos(x)=cos(x+pi/2)

 

[/math]

 

I know from experience that 7pi/4 and 3pi/4 are the solutions, but is there a way to derive it?

 

 

It can be solved geometrically, If I have time tomorrow I might draw it up real quick and post it if no one else beats me to it. Sorry for leaving it like this but I'm tired.

Link to post
Share on other sites
as stated' date=' is there a none intuitive way of solving

 

[math']

 

cos(x)=cos(x+pi/2)

 

[/math]

 

I know from experience that 7pi/4 and 3pi/4 are the solutions, but is there a way to derive it?

 

Hey,

There is a formula:

[math]cos(A+B) = cos(A)cos(B)-sin(A)sin(B)[/math]

so,

[math]cos(x+\tfrac{\pi}{2}) = cos(x)cos(\tfrac{\pi}{2})-sin(x)sin(\tfrac{\pi}{2})[/math]

[math]cos(x+\tfrac{\pi}{2}) = - sin(x)[/math]

 

so your problem becomes

[math]cos(x) = sin(x)[/math]

Lemma:[math]cos(x) =/= 0[/math]

because

if [math]cos(x) = 0[/math],

then [math] sin(x) =/= 0[/math]

 

So,

[math]cos(x)\frac{1}{cos(x)} = sin(x)\frac{1}{cos(x)}[/math]

[math]tan(x) = 1[/math]

this happens at

[math]x = {x| x = \tfrac{\pi}{4} + k\pi} [/math] For all integer [math]k[/math]

 

Does that help?

Link to post
Share on other sites
as stated' date=' is there a none intuitive way of solving

 

[math']

 

cos(x)=cos(x+pi/2)

 

[/math]

 

I know from experience that 7pi/4 and 3pi/4 are the solutions, but is there a way to derive it?

What do you mean with non-intuive way? Two cosines are the same when either the arguments are the same or opposites, with 2k*pi of course.

Link to post
Share on other sites

EDIT: I misunderstood the question, I´m afraid. So the first part that was here is skipped, now.

 

Generally, relations involving cosines and sines often become trivial if one rewrites then as the linear combination of exp(ix) and exp(-ix).

Link to post
Share on other sites
Perhaps it would help if rakuenso would edit out his typo. I don´t really know what the question is about but it´s most certainly not about "cos(x) = cos(x + pi/2)" since disproving this one is trivial: Let x=0 => cos(x)=cos(0)=1' date=' cos(x+pi/2)=cos(pi/2)=0, 1 != 0.

 

Generally, relations involving cosines and sines often become trivial if one rewrites then as the linear combination of exp(ix) and exp(-ix).[/quote']

I think the proposed problem was an equation ("solve for x") and not an identity ("proof for all x").

Link to post
Share on other sites

Ok :)

 

Well in general for equations like this, we have that

 

[math]\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + 2k\pi \vee \alpha = - \beta + 2k\pi ,\forall k \in \mathbb{Z}[/math]

 

I don't consider this as "intuitive" since this holds in general and is a perfect algebraic solution - but perhaps the topic starter can clarify what he meant.

Link to post
Share on other sites
Hey' date='

There is a formula:

[math']cos(A+B) = cos(A)cos(B)-sin(A)sin(B)[/math]

so,

[math]cos(x+\tfrac{\pi}{2}) = cos(x)cos(\tfrac{\pi}{2})-sin(x)sin(\tfrac{\pi}{2})[/math]

[math]cos(x+\tfrac{\pi}{2}) = - sin(x)[/math]

 

...

Lemma:[math]cos(x) =/= 0[/math]

because

if [math]cos(x) = 0[/math],

then [math] sin(x) =/= 0[/math]

 

...

 

Does that help?

 

Wow, I feel stupid.

 

After the cosine formula manipulation, the problem becomes:

[math]cos(x) = -sin(x)[/math]

So,

[math]cos(x)\frac{1}{cos(x)} = -sin(x)\frac{1}{cos(x)}[/math]

[math]tan(x) = -1[/math]

this happens at

[math]x = \tfrac{3\pi}{4} + k\pi [/math] For all integer [math]k[/math]

 

There, that should work.

 

P.S. So the first few positive x that will satisfy the equation are [math]\tfrac{3\pi}{4}, \tfrac{7\pi}{4}, \tfrac{11\pi}{4}, \tfrac{15\pi}{4}, \tfrac{19\pi}{4},...[/math]

Link to post
Share on other sites
Ok :)

 

Well in general for equations like this' date=' we have that

 

[math']\cos \alpha = \cos \beta \Leftrightarrow \alpha = \beta + 2k\pi \vee \alpha = - \beta + 2k\pi ,\forall k \in \mathbb{Z}[/math]

 

 

You can show this as follows:

 

[math]cos u = cos v \Leftrightarrow cos u - cos v = -2 sin\frac{u - v}{2} sin \frac{u+v}{2} = 0[/MATH].

 

[MATH]sin\frac{u - v}{2} = 0 \Leftrightarrow u - v = 2\pi n[/MATH]

 

[MATH]sin\frac{u + v}{2} = 0 \Leftrightarrow u + v = 2\pi n[/MATH].

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.