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Trig problem

  1. Jul 29, 2008 #1
    OK, first, I'm a grown man, this is not a homework problem. Well, in a way it is, since I'm teaching myself, but anyway.

    I have finally graduated to trig and the six basic functions and how they relate. What I'm having a problem with is this:

    sin/1+cos = 1-cos/sin

    How is this true? I've done every algebraic manipulation I can think of, but I can't see how this can be. I'll show what I have come up with, using the fact that sin x =y/r and cos x =x/r.

    It should break down like this: (y/r)/1+(x/r) = [1-(x/r)]/(y/r)

    Now, here's how I approached it: Left of the equals sign (y/r)/1+(x/r) = y/(r+x)
    Then right of the equals sign: [1-(x/r)]/(y/r) = (r-x)/y

    So when does y/(r+x) ever equal (r-x)/y? Thanks for any help provided. And I hope this wasn't too simple for you guys to consider answering.
     
  2. jcsd
  3. Jul 29, 2008 #2

    HallsofIvy

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    Do you mean sin x/(1+ cos x)= (1- cos x)/sin x? What you wrote would be, strictly, interpreted as (sin x/1)+ cos x= 1- (cos x/sin x) which not true! Parentheses are important and, please, never write a function without a variable!

    If you multiply both sides of the equation by sin x(1+ cos x), to get rid of the fractions, you get sin2 x= (1+ cos x)(1- cos x)= 1- cos2 x.
    Does that remind you of a basic identity?

    To answer your last question y/(r+x)= (r- x)/y if and only if (multiplying both sides by y(r+x)) y2= r2- x2, which is the same as x2+ y2= r2.
     
  4. Jul 29, 2008 #3
    Well, duh! That makes perfect sense! For some reason, it never occurred to me to treat it like any other fraction and find the LCD. Of course, I intentionally omitted the (x), assuming we all knew it would be "sin(x)" and "cos(x)". Sorry. That's how I've gotten used to doing it when I scribble. Thanks, I'm sure you can tell I'm still struggling with the basic trig identities, but I must say, they are vastly easier than I remember them being in High School.
     
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