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By using inverse trig identities, it can be shown that $sec(x)*cos(x)=1$. However, when $x=π/2$, the resulting function would be $$(1/0)*(0/1)=1,$$ and cross multiplying yields $$1*0=1.$$ I'm not sure where I went wrong in this example but there doesn't seem to be any apparent gaps in the graph of $sec(x)*cos(x)$. What am I doing wrong?

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    $\begingroup$ $\sec(\pi/2)$ is not defined. $\endgroup$ – sinbadh Jan 4 '16 at 16:42
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    $\begingroup$ The problem is that it is division by zero. You can not divide by zero and expect good things to happen. $\endgroup$ – JMoravitz Jan 4 '16 at 16:45
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    $\begingroup$ It cannot be defined as $\frac10$, since that's not a valid fraction. $\endgroup$ – Arthur Jan 4 '16 at 16:45
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    $\begingroup$ Why is this getting downvoted? Though naïve, this is a perfectly valid question, and the OP has clearly demonstrated that he/she has put thought into it. $\endgroup$ – wgrenard Jan 4 '16 at 17:43
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    $\begingroup$ It shouldn't be downvoted and we shouldn't insult naive people. I probably shouldn't yell at them either. But I do get emphatic about basic points. $\endgroup$ – fleablood Jan 4 '16 at 17:48
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  1. The domain of $\sec x$ does not include $\frac{\pi}{2}$; $\sec\frac{\pi}{2}$ is not defined. More information here.

  2. Even if you didn't know that, or forgot it, $\frac{1}{0}$ is not defined.

  3. Even if you forgot those things, $0\cdot\frac{1}{0} = \frac{0}{0}$ by your logic, which is undefined.

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The terms $1/0,0/0,\infty$ are indeterminate or undefined and cant be used for algebra eg $6\times 0=5\times 0$ cancelling zero we get $6=5$ so you get why they cant be used !

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  • $\begingroup$ @cr3 i have edited $\endgroup$ – Archis Welankar Jan 4 '16 at 16:59
  • $\begingroup$ @ArchisWelankar Its nice . $\endgroup$ – Tesla Jan 4 '16 at 16:59
  • $\begingroup$ $\infty$ by itself is not indeterminate nor undefined, but you're right that it can't be used for algebra because it's not a number. $\endgroup$ – cr3 Jan 4 '16 at 17:01
  • $\begingroup$ Thanks @ Tesla and yes that was my motive algebraically useless numbers @ cr3 $\endgroup$ – Archis Welankar Jan 4 '16 at 17:06
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    $\begingroup$ You could just say "Undefined terms like $1/0$, indeterminate terms like $0/0$ and not actual numbers like $\infty$ can't be used..." rather than mislabel $\infty$ as indeterminate or undefined. $\endgroup$ – cr3 Jan 4 '16 at 17:10
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$m$ divided by $n$ is the specific number, $q$, such that $qn = m$. So $m$ divided by $0$ is the specific number, $q$, such that $q*0 = m$. There is no such number so division by 0 is impossible and 1/0 is undefined.

Now, trigonometry. You have a unit circle. You have an angle $\theta$. The points on the unit circle corresponding to the angle $\theta$ are $(x, y) = (\cos \theta, \sin \theta)$ and there is an inscribed right triangle corresponding to that angle with sides $x = \cos \theta$, $y = \sin \theta$ and hypotenuse = 1. That's basic.

So... Draw a tangent line from the (x,y) point that is tangent to the circle. Extend that line until it crosses the x-axis (!!Warning!! !!Warning!! Who ever told you it would cross the x-axis?) (Oh, shut up, voice in my head; most lines will cross the x-axis.)... extend that line until it crosses the x-axis. This forms another right angle that is similar to the first. The tangent line from $(x,y)$ to the x-axis coresponds to the $x = \cos \theta$ side of the first triangle. The radius of the unit circle (from the origin to $(x,y)$) on the new triangle corresponds with the $y = \sin \theta$ side of the original triangle. The x-axis from the origin to where the line intersects the x-axis corresponds to the hypotenuse of the original triangle.

Doing a bit of geometry with realize the scaling factor of this new triangle is $1/\cos \theta$. So the distance of the tangent point to the x-axis is $\sin \theta /\cos \theta$. We call that $\tan \theta$ (because it's the length of a tangent chord!!!). And the length from the $(0,0)$ origin to where the tangent line intersects the x-axis is $1/\cos \theta$. We call that $\sec \theta$ (because it is the length of a secant chord !... wait, what the heck is a "secant chord"? .. ).

Let me repeat that as I'm sure your eyes are glazing over.

$\sec \theta$ is the distance from the origin to where the tangent line intersects the x-axis.

Okay, so let $\theta = \pi/2$. $\sec \pi/2$ is the distance from the origin to where the tangent line intersects the x-axis. But the tangent line is parallel to the x-axis!!!! The tangent line never intersects the x-axis and there is no such distance!

(Maybe I should have listened to that voice in my head. Wouldn't be the first time. Of course, there was the time when it told me to... well, never mind.)

So $\sec \pi/2$ is undefined.

Just like $1/0$ is undefined.

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There are many similarities. $\sec \theta = 1/\cos \theta$ so $\sec \pi/2$ would $= 1/0$ if such a thing were possible.

Also notice as $\theta$ gets closer and closer to $\pi/2$ the tangent line gets larger and larger and $\sec \theta$ gets larger and larger and approaches infinity.

Likewise if you are dividing $m$ by $n$ and $n$ gets smaller and smaller and closer to 0 then $m/n$ gets larger and approaches infinity.

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