Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Trigonometry Proof


Date: 7/18/96 at 4:32:59
From: Anonymous
Subject: Trig. Proof

I would appreciate any help that you may be able to give.

 (cotX)(cosX)          cosX
--------------     =  -------
(cotX) + (cosX)       1+ sinX        
		         

I have to prove this question, not answer or solve.

Thank you in advance.


Date: 7/18/96 at 13:1:55
From: Doctor Paul
Subject: Re: Trig. Proof

Here's how to prove both sides of your equation are equal:

cot(x) * cos(x)           cos(x)
---------------    =    ---------
cot(x) + cos(x)         1 + sin(x)


I really didn't know where to begin.  There is usually some sort of
substitution to make in these trig identity problems 
(like (sin(x))^2 = 1 - (cos(x))^2) but there weren't any obvious
substitutions here.  I began by cross multiplying:

[cot(x)*cos(x)]+[cot(x)*cos(x)*sin(x)] = [cot(x)*cos(x)] + [(cos(x)*
cos(x)]

Recall that tan(x) is really sin(x) / cos(x)
and that cot(x) is the 1 / tan(x), so cot(x) = cos(x) / sin(x).

Let's make that substitution:

[cos(x)*cos(x)]    [cos(x)*cos(x)*sin(x)]    
 -------------   +  ---------------------  =
     sin(x)            sin(x)              


[cos(x)*cos(x)]     [(cos(x)*cos(x)]
 -------------    +  ------------- 
    sin(x)                1

I don't think it's too hard to see the two sides are equal..just 
cancel the two sin(x) terms in the second of the four brackets and 
you're in business..

-Doctor Paul,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 7/18/96 at 15:59:8
From: Doctor Anthony
Subject: Re: Trig. Proof

>prove (cot x * cos x)/(cot x + cos x) = (cos x)/(1 + sin x)

These questions are easily answered if you remember the relationships 
between the various trig. ratios.  In this case use  
cot(x) = cos(x)/sin(x)

So the lefthand side of the above expression can be written:

  [cos(x)/sin(x)]*cos(x)
  ----------------------
  cos(x)/sin(x) + cos (x)


Multiplying the top and bottom by sin(x), we get 

    cos(x)*cos(x)
  ---------------------
   cos(x) + sin(x)*cos(x)

Dividing the top and bottom by cos(x)  and we get:

    cos(x)
   ---------   = righthand side.
   1 + sin(x)


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/