as stated, 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?

I don't think you need to derive it. Can't you just use the cosine curve?

 

oh.. i counted that as an intuitive approach

oh.. i counted that as an intuitive approach

ah, I see what you mean. Then either it can't be done or I don't know how to do it. The latter is more likely.

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.

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?

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.

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).

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").

Yep, see my edit

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.

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]

ok thx =)

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].

ok thx =)

 

your welcome!