rakuenso 11 Posted November 16, 2005 Share Posted November 16, 2005 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? Link to post Share on other sites

ecoli 372 Posted November 16, 2005 Share Posted November 16, 2005 I don't think you need to derive it. Can't you just use the cosine curve? Link to post Share on other sites

rakuenso 11 Posted November 16, 2005 Author Share Posted November 16, 2005 oh.. i counted that as an intuitive approach Link to post Share on other sites

ecoli 372 Posted November 16, 2005 Share Posted November 16, 2005 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. Link to post Share on other sites

jcarlson 17 Posted November 16, 2005 Share Posted November 16, 2005 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

cosine 10 Posted November 16, 2005 Share Posted November 16, 2005 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

TD 10 Posted November 16, 2005 Share Posted November 16, 2005 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

timo 578 Posted November 16, 2005 Share Posted November 16, 2005 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

TD 10 Posted November 16, 2005 Share Posted November 16, 2005 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

timo 578 Posted November 16, 2005 Share Posted November 16, 2005 Yep, see my edit Link to post Share on other sites

TD 10 Posted November 16, 2005 Share Posted November 16, 2005 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

cosine 10 Posted November 16, 2005 Share Posted November 16, 2005 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

rakuenso 11 Posted November 16, 2005 Author Share Posted November 16, 2005 ok thx =) Link to post Share on other sites

Algebracus 10 Posted November 18, 2005 Share Posted November 18, 2005 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

cosine 10 Posted November 18, 2005 Share Posted November 18, 2005 ok thx =) your welcome! Link to post Share on other sites

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