Exploring the Truth Behind: sin/1+cos = 1-cos/sin

[DecayProduct]

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.

 

[HallsofIvy]

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.

 

[DecayProduct]

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.