My Math Forum Need help with some alien trigonometry
 User Name Remember Me? Password

 New Users Post up here and introduce yourself!

September 20th, 2017, 04:08 PM   #1
Newbie

Joined: Sep 2017
From: Lahore

Posts: 4
Thanks: 0

Need help with some alien trigonometry

Show that y≈∆φ×secφ in the jpeg attached.

or ∆y= sec φ

A and B are points on curved surface, two lines are extended through origin to a line that is tangent to the circle, these points are A' and B', change in Angle will bring a change in length between A' and B'. I need to know how is this do-able.
Attached Images
 maths (2).jpg (9.6 KB, 9 views)

Last edited by Wasif Jalal; September 20th, 2017 at 04:30 PM. Reason: error

 September 20th, 2017, 04:47 PM #2 Senior Member   Joined: Aug 2012 Posts: 1,661 Thanks: 427 Looks doable. One place to start is that $B'$ and $A'$are $\tan \varphi$ and $\tan(\varphi + \Delta \varphi)$, respectively. That's a little known factoid but it's easy to see why it's true. And it gives insight into why it's called the tangent. Also note that if the coordinates of $B$ are $(\cos \varphi, \sin \varphi)$ then the coordinates of $B'$ are $(|B| \cos \varphi, |B| \sin \varphi)$ where $|B|$ is the length of $B$. [I'm taking the circle to be the unit circle]. Those are some of the trig relationships I'd start playing around with. And eventually you have to get it in the form of a secant using some trig identity. ps -- Also I'd play around with the addition formula since they relate the coordinates of $A$ and $B$ to each other. By the way one of your relationships is approximate equality and the other is exact equality. Is that what you intend to say? If the relationship is only approximately equal you have to think about it differently. Last edited by Maschke; September 20th, 2017 at 05:39 PM.
 September 20th, 2017, 05:59 PM #3 Senior Member   Joined: Aug 2012 Posts: 1,661 Thanks: 427 That second one doesn't look right. $\sec \theta = \frac{1}{\cos \theta}$. So when you say $\Delta y = \sec \varphi = \frac{1}{\cos \varphi}$ you're saying that $\Delta y$ doesn't depend on $\Delta \varphi$. There must must some more info or there's a typo or perhaps I'm not reading the diagram accurately. Last edited by Maschke; September 20th, 2017 at 06:02 PM.
 September 20th, 2017, 08:21 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,661 Thanks: 965 Math Focus: Elementary mathematics and beyond This problem is posted on another forum. Similar questions were raised there.
 September 20th, 2017, 10:56 PM #5 Newbie   Joined: Sep 2017 From: Lahore Posts: 4 Thanks: 0 replying to: greg1313 It was I who posted the same question. Last edited by skipjack; September 21st, 2017 at 01:07 AM.
 September 20th, 2017, 11:04 PM #6 Newbie   Joined: Sep 2017 From: Lahore Posts: 4 Thanks: 0 What relationships do you find equal in the problem? Also, please could you me further with this problem; kinda desperate. Last edited by skipjack; September 21st, 2017 at 01:07 AM.
 September 21st, 2017, 01:55 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,247 Thanks: 1439 $\Delta y = \tan(\varphi + \Delta\varphi) - \tan(\varphi) = \tan(\Delta\varphi)(1 + \tan(\varphi + \Delta\varphi)\tan(\varphi))$ If $\Delta\varphi$ is very small, $\tan(\Delta\varphi) \approx \Delta\varphi$ and $\tan(\varphi + \Delta\varphi) \approx \tan(\varphi)$, so $\Delta y \approx \Delta\varphi(1 + \tan^2(\varphi) = \Delta\varphi\sec^2(\varphi)$. Thanks from Maschke

 Tags alien, trigonometry

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post caters Algebra 9 March 6th, 2014 12:40 PM John Marsh Trigonometry 3 September 6th, 2013 03:27 AM johnny Algebra 6 March 7th, 2011 02:06 PM johnny Algebra 7 March 5th, 2011 09:39 AM Trigeometry 93 New Users 1 April 16th, 2009 08:18 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top