The parametric equation of cardioid is $$(x(t),y(t))=(a(2\cos t-\cos 2t), a(2\sin t-\sin 2t)).$$ How To underdstand from parametric equation that this curve is symmetric about $x$-axis?
Can anyone explain that in detail?
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Hint: $x$ is an even function of $t$ and $y$ an odd one. – Sep 28 '17 at 09:19
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@YvesDaoust, Yes it is true that $x(-t)=x(t)$ and $y(-t)=-y(t)$. But how it can help to my question? – RFZ Sep 28 '17 at 09:23
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What is symmetry about $x$ axis ? – Sep 28 '17 at 09:24
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I meant that the graph of function is symmetric with regard to $x$-axis, i.e. $(x,y)$ lies in graph then $(x,-y)$ also lies. – RFZ Sep 28 '17 at 09:27
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Now merge your last two comments. – Sep 28 '17 at 09:27
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@YvesDaoust, if $(x(t),y(t))\in \Gamma$ and $x(-t)=x(t), \ y(-t)=-y(t)$. Then: $(x(t),y(t))=(x(-t), -y(-t))\in \Gamma$. But how to derive that $(x(t), -y(t)) \in \Gamma$? – RFZ Sep 28 '17 at 09:37
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It is in front of your eyes. Try harder. – Sep 28 '17 at 09:39
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@YvesDaoust Please can you tell me whether my method is even applicable here.? – jonsno Sep 28 '17 at 09:46
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@samjoe: approximately because $x$ may not be a univocal function of $y$. But the question is so simple (!) – Sep 28 '17 at 09:56
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@YvesDaoust, Unfortunately my efforts are worthless. I am not able to show that $(x(t),-y(t))$ also $\in \Gamma$ – RFZ Sep 28 '17 at 10:01
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@RFZ Put $-t$ in place of $t$ ! – jonsno Sep 28 '17 at 10:04
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@samjoe, suppose $f(t)=(x(t),y(t))$ lies on graph and if we put $-t$ instead of $t$ we get $f(-t)=(x(-t),y(-t))=(x(t), -y(t))$ (due to parity of functions). But why $f(-t)$ also lies on graph? Maybe it is simple but I can't comprehend it. – RFZ Sep 28 '17 at 10:08
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@RFZ You have imposed no condition on $t$, so it can be any real number! – jonsno Sep 28 '17 at 10:11
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If I am not mistaken $t$ is any real so $f(-t)$ also lies on graph and hence it's symmetrical about $x$-axis. Right? – RFZ Sep 28 '17 at 10:13
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@RFZ Yes thats all! – jonsno Sep 28 '17 at 10:17