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

[rakuenso]

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?

[ecoli]

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

 

[rakuenso]

oh.. i counted that as an intuitive approach

[ecoli]

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.

[jcarlson]

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.

[cosine]

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?

[TD]

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.

[timo]

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).

[TD]

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").

[timo]

Yep, see my edit

[TD]

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.

[cosine]

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]

[rakuenso]

ok thx =)

[Algebracus]

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].

[cosine]

ok thx =)

 

your welcome!