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anonymous

  • one year ago

HELP NEEDED!!! WILL FAN AND MEDAL!!!! I need help understanding how to prove identities.

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  1. anonymous
    • one year ago
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    I have a specific example. Please help...

  2. anonymous
    • one year ago
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    what is it?

  3. anonymous
    • one year ago
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    \[\tan \frac{ \cos \theta }{ 1+ \sin \theta }=\frac{ 1 }{ \cos \theta }\]

  4. anonymous
    • one year ago
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    It was actually supposed to be tangent theta times cosine theta over one plus sine theta

  5. anonymous
    • one year ago
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    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.

  6. anonymous
    • one year ago
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    \[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] this is what I need help combining because I have already converted Tan

  7. anonymous
    • one year ago
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    what the heck is the tan doinog there ?

  8. anonymous
    • one year ago
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    The tan was there because our teacher was trying to get us to recreate all of the identities to where they equal one another

  9. anonymous
    • one year ago
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    no don't convert to tangent in fact, if it was tangent, you would convert to \(\frac{\sin(x)}{\cos(x)}\)

  10. anonymous
    • one year ago
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    lets do a tiny bit of algebra first

  11. anonymous
    • one year ago
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    That's what I did and that was my farthest point in this.

  12. anonymous
    • one year ago
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    ok i got that wrong, hold the phone

  13. anonymous
    • one year ago
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    \[\frac{b}{a}+\frac{a}{1+b}\] that's better can you add these?

  14. anonymous
    • one year ago
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    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

  15. anonymous
    • one year ago
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    lol that is always the problem, the algebra

  16. anonymous
    • one year ago
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    here is the one true way to add fractions \[\huge \frac{A}{B}+\frac{C}{D}=\frac{AD+BC}{BD}\]

  17. anonymous
    • one year ago
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    o.o"

  18. anonymous
    • one year ago
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    in your case you will have \[\frac{b(1+b)+a^2}{a(1+b)}\]

  19. anonymous
    • one year ago
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    I am getting more lost by the minute...

  20. anonymous
    • one year ago
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    that is the way to add fractions if they are numbers\[\frac{2}{7}+\frac{3}{5}=\frac{2\times 5+7\times 3}{7\times 5}\]

  21. anonymous
    • one year ago
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    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

  22. anonymous
    • one year ago
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    ohhhh ok, I get it, sort of.

  23. anonymous
    • one year ago
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    forget that "least common denominator" nonsense you were taught that is the way to add, you cannot avoid it

  24. anonymous
    • one year ago
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    so ready to start again?

  25. anonymous
    • one year ago
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    Yeah I suppose

  26. anonymous
    • one year ago
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    using the one true way to add \[\frac{b}{a}+\frac{a}{1+b}\] you get \[\frac{b(1+b)+a^2}{a(1+b)}\]

  27. anonymous
    • one year ago
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    now some more algebra, but just a little \[\frac{b(1+b)+a^2}{a(1+b)}=\frac{b+b^2+a^2}{a(1+b)}\]

  28. anonymous
    • one year ago
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    that is just the distributive law in the numerator now that we are done with algebra, we can go back to sines and cosines

  29. anonymous
    • one year ago
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    sooooooo then \[\frac{ \cos \theta }{ \sin \theta } + \frac{ \cos \theta }{ 1+\sin \theta }\] should become....

  30. anonymous
    • one year ago
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    \[\frac{\sin(x)+\sin^2(x)+\cos^2(x)}{\cos(x)(1+\sin(x))}\]

  31. anonymous
    • one year ago
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    Ok. But now comes an even trickier part.... I have to find a way to reduce that

  32. anonymous
    • one year ago
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    hold a sec the initial question was \[\frac{ \sin \theta }{ \cos \theta }+\frac{ \cos \theta }{ 1+ \sin \theta }\] right?

  33. anonymous
    • one year ago
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    Yes

  34. anonymous
    • one year ago
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    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

  35. anonymous
    • one year ago
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    the usual miracle occurs, \[\sin^2(x)+\cos^2(x)=1\] the mother of all trig identities

  36. anonymous
    • one year ago
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    correct

  37. chrisplusian
    • one year ago
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    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

  38. anonymous
    • one year ago
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    \[\frac{\sin(x)+\overbrace{\sin^2(x)+\cos^2(x)}^{\text{ this is 1}}}{\cos(x)(1+\sin(x))}\]

  39. anonymous
    • one year ago
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    \[\frac{\sin(x)+1}{\cos(x)(1+\sin(x))}\] NOW you can cancel the common factor of \[1+\sin(x)\] top and bottom

  40. anonymous
    • one year ago
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    Ok, ok, I see it now. so the Sins actually match and leave you with \[\frac{ 1 }{ \cos \theta }\]

  41. anonymous
    • one year ago
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    yes if you want to impress your teacher, write it as \[\sec(\theta)\]

  42. anonymous
    • one year ago
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    i hope you also see that 98% of this is algebra

  43. anonymous
    • one year ago
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    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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