anonymous
  • anonymous
HELP NEEDED!!! WILL FAN AND MEDAL!!!! I need help understanding how to prove identities.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
I have a specific example. Please help...
anonymous
  • anonymous
what is it?
anonymous
  • anonymous
\[\tan \frac{ \cos \theta }{ 1+ \sin \theta }=\frac{ 1 }{ \cos \theta }\]

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anonymous
  • anonymous
It was actually supposed to be tangent theta times cosine theta over one plus sine theta
anonymous
  • anonymous
I have all of the identities written out so I can easily refer to them, but the problem I am having is learning how to combine the like terms.
anonymous
  • anonymous
\[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] this is what I need help combining because I have already converted Tan
anonymous
  • anonymous
what the heck is the tan doinog there ?
anonymous
  • anonymous
The tan was there because our teacher was trying to get us to recreate all of the identities to where they equal one another
anonymous
  • anonymous
no don't convert to tangent in fact, if it was tangent, you would convert to \(\frac{\sin(x)}{\cos(x)}\)
anonymous
  • anonymous
lets do a tiny bit of algebra first
anonymous
  • anonymous
That's what I did and that was my farthest point in this.
anonymous
  • anonymous
ok i got that wrong, hold the phone
anonymous
  • anonymous
\[\frac{b}{a}+\frac{a}{1+b}\] that's better can you add these?
anonymous
  • anonymous
I have no idea how to... or at least I can't remember. I know with multiplying you foil it, if you divide you use the kcf format :/ but adding and subtracting makes no sense to me because all I know is a common denominator is required
anonymous
  • anonymous
lol that is always the problem, the algebra
anonymous
  • anonymous
here is the one true way to add fractions \[\huge \frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}\]
anonymous
  • anonymous
o.o"
anonymous
  • anonymous
in your case you will have \[\frac{b(1+b)+a^2}{a(1+b)}\]
anonymous
  • anonymous
I am getting more lost by the minute...
anonymous
  • anonymous
that is the way to add fractions if they are numbers\[\frac{2}{7}+\frac{3}{5}=\frac{2\times 5+7\times 3}{7\times 5}\]
anonymous
  • anonymous
and it is also the way to add fractions if they have variables it is the only way to do this, by using algebra to add
anonymous
  • anonymous
ohhhh ok, I get it, sort of.
anonymous
  • anonymous
forget that "least common denominator" nonsense you were taught that is the way to add, you cannot avoid it
anonymous
  • anonymous
so ready to start again?
anonymous
  • anonymous
Yeah I suppose
anonymous
  • anonymous
using the one true way to add \[\frac{b}{a}+\frac{a}{1+b}\] you get \[\frac{b(1+b)+a^2}{a(1+b)}\]
anonymous
  • anonymous
now some more algebra, but just a little \[\frac{b(1+b)+a^2}{a(1+b)}=\frac{b+b^2+a^2}{a(1+b)}\]
anonymous
  • anonymous
that is just the distributive law in the numerator now that we are done with algebra, we can go back to sines and cosines
anonymous
  • anonymous
sooooooo then \[\frac{ \cos \theta }{ \sin \theta } + \frac{ \cos \theta }{ 1+\sin \theta }\] should become....
anonymous
  • anonymous
\[\frac{\sin(x)+\sin^2(x)+\cos^2(x)}{\cos(x)(1+\sin(x))}\]
anonymous
  • anonymous
Ok. But now comes an even trickier part.... I have to find a way to reduce that
anonymous
  • anonymous
hold a sec the initial question was \[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
so after the algebra we get to \[\frac{\sin(x)+\sin^2(x)+\cos^2(x)}{\cos(x)(1+\sin(x))}\] don't look to reduce yet
anonymous
  • anonymous
the usual miracle occurs, \[\sin^2(x)+\cos^2(x)=1\] the mother of all trig identities
anonymous
  • anonymous
correct
chrisplusian
  • chrisplusian
The reasoning behind adding and subtracting the fractions is exactly what was said above, but the step being left out from above that might make it clear is that you just multiply both fractions by a number that will make both denominators the same or "common". In the above example....\[\frac{ b }{ a }+\frac{ a }{ 1+b } = (\frac{ 1+b }{ 1+b })(\frac{ b }{ a })+(\frac{ a }{ a })(\frac{ a }{ 1+b })\] and that is where the other equation @satellite73 came up with came from. Not to inturrupt
anonymous
  • anonymous
\[\frac{\sin(x)+\overbrace{\sin^2(x)+\cos^2(x)}^{\text{ this is 1}}}{\cos(x)(1+\sin(x))}\]
anonymous
  • anonymous
\[\frac{\sin(x)+1}{\cos(x)(1+\sin(x))}\] NOW you can cancel the common factor of \[1+\sin(x)\] top and bottom
anonymous
  • anonymous
Ok, ok, I see it now. so the Sins actually match and leave you with \[\frac{ 1 }{ \cos \theta }\]
anonymous
  • anonymous
yes if you want to impress your teacher, write it as \[\sec(\theta)\]
anonymous
  • anonymous
i hope you also see that 98% of this is algebra
anonymous
  • anonymous
which, if your algebra is not what it needs to be, is going to be a problem, so bone up on it also use the gimmick of replacing sine and cosine by letters to make the algebra easier if we had to do this writing sine and cosine each time it would have been a pain

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