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Math Help - Need help with another identity...

  1. #1
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    Need help with another identity...

    I have no idea what to do with this one...:

    sin x + cos x = (sin x / (1 - cos x/sin x)) + (cos x /1 - sin x/cos x))

    Help please... this assignment is due tomorrow and I'm not even half way done...
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  2. #2
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    ok - start on the right side - I would play with it - start by changing it to \frac{cosx}{sinx} to maybe cotx as well as the other to tanx
    then find your identity chart thing and see if theres something you can do with that
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  3. #3
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    Well there isn't. I already checked that. That was really obvious.

    But thanks for trying though.
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  4. #4
    Member cassiopeia1289's Avatar
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    lol, ok, work with what you got though -sometimes it helps (although you should only work on one side) to rewrite the other side in different forms - you can see different angles that which you might be able to approach the final one
    I'm reluctant to give you the flat out answer, but the really only advise with identities is to play with them - try different things and just chug at it
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  5. #5
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    Well thats what I've been doing all day, for the past 6-7 hours at least, and I have 5 problems done.

    I really am stumped on this one...
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  6. #6
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    I'm guessing your problem is this one. Please try to be less ambiguous when typing up the problems:

    \sin x + \cos x = \frac{\sin x}{1 - \frac{\cos x}{\sin x}} + \frac{\cos x}{1 - \frac{\sin x}{\cos x}}

    Combine the denominators into single fractions:

    = \frac{\sin x}{{\color{red} \frac{\sin x}{\sin x}} - \frac{\cos x}{\sin x}} + \frac{\cos x}{{\color{blue}\frac{\cos x}{\cos x}} - \frac{\sin x}{\cos x}}

    = \frac{\sin x}{\frac{\sin x - \cos x}{\sin x}} + \frac{\cos x}{\frac{\cos x - \sin x}{\cos x}}

    = \frac{\sin^{2} x}{\sin x - \cos x} + \frac{\cos^{2} x}{\cos x - \sin x}

    Note that in the denominator of the second fraction, we can factor out a -1 and it'll be the same as the first fraction:
    = \frac{\sin^{2} x}{\sin x - \cos x} + \frac{\cos^{2} x}{-(\sin x - \cos x)}

    So we get:
    = \frac{\sin^{2} x}{\sin x - \cos x} {\color{red}-} \frac{\cos^{2} x}{\sin x - \cos x}

    Now combine the fractions and factor out the numerator (hint: difference of squares)
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  7. #7
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    Okay... that all makes sense, but I dont know how to do difference of squares on a fraction like that... you're over estimating me... please show me.
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  8. #8
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    It's not too bad. Recall the difference of squares:
    {\color{red} a^{2}} - {\color{blue}b^{2}} = ({\color{red}a} - {\color{blue}b})({\color{red}a} + {\color{blue}b})

    Now for the numerator of your expression, imagine:
    {\color{red} a^{2} = \sin^{2} x} and {\color{blue} b^{2} = \cos^{2} x}
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  9. #9
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    Okay, that gives me...

    sin x + cos x/ sin x - cos x

    Since the other one is equal to one, it leaves me with this...
    Now what can I do? I'm so confused...

    (And sorry about the formatting of my math text, I don't know how to do fraction bars and stuff)
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  10. #10
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    No ... one shouldn't disappear.

    \frac{\sin^{2}x - \cos^{2} x}{\sin x - \cos x} = \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x}

    Does something cancel?

    The identity you're referring to is: \sin^{2} x + \cos^{2} x = 1 (The sin and cos are squared![/tex]
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  11. #11
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    OH!

    I seperated the two... like this:

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

    But I get it now.

    Thanks!

    Well thats 6 done, about 17 more to go...
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