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September 20th, 2017, 03:08 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 doable. Last edited by Wasif Jalal; September 20th, 2017 at 03:30 PM. Reason: error 
September 20th, 2017, 03:47 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,999 Thanks: 573 
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 04:39 PM. 
September 20th, 2017, 04:59 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,999 Thanks: 573 
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 05:02 PM. 
September 20th, 2017, 07:21 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,854 Thanks: 1078 Math Focus: Elementary mathematics and beyond 
This problem is posted on another forum. Similar questions were raised there.

September 20th, 2017, 09:56 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 12:07 AM. 
September 20th, 2017, 10:04 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 12:07 AM. 
September 21st, 2017, 12:55 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,515 Thanks: 1745 
$\Delta y = \tan(\varphi + \Delta\varphi)  \tan(\varphi) = \tan(\Delta\varphi)(1 + \tan(\varphi + \Delta\varphi)\tan(\varphi))$ If $\Delta\varphi$ is very small, $\tan(\Delta\varphi) \approx \Delta\varphi$ and $\tan(\varphi + \Delta\varphi) \approx \tan(\varphi)$, so $\Delta y \approx \Delta\varphi(1 + \tan^2(\varphi) = \Delta\varphi\sec^2(\varphi)$. 

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