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I am confused about what is the geometric representation and interpretation of the secant and cosecant of an angle. I understand how to calculate them but I do not know what they mean, geometrically.

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  • $\begingroup$ I doubt there is really any.. they're just there to help you later on in future math courses; e.g. the derivative (calculus term) of the tangent function is secant squared. $\endgroup$
    – y_prime
    Dec 12, 2016 at 23:48
  • $\begingroup$ @pie314271 Do you still have doubts after seeing the diagrams below? $\endgroup$
    – Théophile
    Dec 13, 2016 at 0:16
  • $\begingroup$ @Théophile: I was thinking that he was referring to the usages of secant/cosecant (e.g. $e^{ix}=\cos x+i\sin x$). Of course there's that, but based on the OP's response you're probably right. $\endgroup$
    – y_prime
    Dec 13, 2016 at 0:34

4 Answers 4

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In the usual terms or geometric representation of cos and sin on the unit circle in terms of some angle $\theta$ you can also get a 'geometric representation' of sec and cosec here also. See the image below.

enter image description here

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  • $\begingroup$ I think the way of looking at it that is shown in Wikipedia's article on trigonometric functions is better. See my posted answer. $\endgroup$
    – Michael Hardy
    Dec 13, 2016 at 0:07
  • $\begingroup$ @MichaelHardy Why do prefer the other diagram? $\endgroup$
    – Théophile
    Dec 13, 2016 at 0:15
  • $\begingroup$ @Théophile : I may actually have to collect my thoughts on this in order to write as short an answer as the question deserves. $\endgroup$
    – Michael Hardy
    Dec 13, 2016 at 1:08
  • $\begingroup$ @Théophile : One reason is that it allows a good explanation of why the tangent and secant are positive in certain quadrants and negative in certain other quadrants. $\qquad$ $\endgroup$
    – Michael Hardy
    Dec 13, 2016 at 1:09
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    $\begingroup$ @MichaelHardy “If I had had more time, I would’ve written a shorter article.” — Mark Twain. $\endgroup$
    – amd
    Dec 13, 2016 at 7:11
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secant cosecan

${{{{{{{{{{{{{{{{{{{{\qquad}}}}}}}}}}}}}}}}}}}}$

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    $\begingroup$ This (rather than the other diagram) is the way I was taught long, long ago. $\endgroup$
    – amd
    Dec 13, 2016 at 7:12
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I am aware that there have been already some answers to this question. However, what helped me really understand the concept was the following interpretation with some narrative, as opposed to just a picture.

  1. Imagine a horizontal and a vertical line crossing
  2. Imagine a unit circle (radius of one) with the centre at that crossing
  3. Draw a line at angle θ from the centre to the circle's boundary and beyond
  4. Draw another vertical line that touches the circle boundary
  5. The distance from the centre of the circle and the point P when the two lines above intercept is the secant of theta

secant_as_a_distant_to_p

This is very well explained at http://www-personal.umich.edu/~copyrght/image/books/Spatial%20Synthesis/trig/ . You can find there an explanation for cosecant as well, among other concepts.

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  • $\begingroup$ This does not add anything new to the question as this is already in Michael Hardy's answer. $\endgroup$
    – soupless
    Nov 24, 2021 at 1:57
  • $\begingroup$ @soupless I understand that from your perspective (the person that already knows the answer) this answer appears not to add anything new. However, it does add more information - namely, the fact that tanΘ line is precisely vertical, or that secΘ is the secant in question, or the link that explains the whole process of arriving at it. I posted my answer from the perspective of a person that did not get the previous answer until they read more about it and optimized it for learning, not for brevity. The link in my answer helped me more that the picture from the answer above. $\endgroup$
    – Michael Szymczak
    Nov 24, 2021 at 22:23
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Sin, cos are circular functions of angle $(\theta)$ they are resolved components of a unit circle as is well known

$$ sin^2 \theta + \cos^2 \theta =1 $$

Using inverse functions definitions as you requested we get

a hyperbola like ( but not a hyperbola) curve plotted on x-, y- axes as shown. The curve does not exist in range/domains $x=\pm1,y=\pm 1.$

enter image description here

It can be parameterized for $ \angle POX= \theta $

$$ a=1, x= a \sec \theta, y= a \csc \theta\;;$$

Although shown, it is rarely used in that form in usage. The labeled circular functions are more in use.

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