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Math Help - Trig Identity - Im so confused!

  1. #1
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    Trig Identity - Im so confused!

    This is my first post. Up until this point I have been able to figure out things by myself but these trig identities are killing me. The ones in my textbook I can figure out but the identities my teacher gives are not as clear to me. Here is the one I am I currently stuck on.

    1-1/2sin(2x)=sin^3(x)+cos^3(x)/sin(x)+cos(x)

    I'm not sure where to start. Perhaps the right side?

    I hope I formatted it right. I don't know how to make it look pretty.

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by repete View Post
    This is my first post. Up until this point I have been able to figure out things by myself but these trig identities are killing me. The ones in my textbook I can figure out but the identities my teacher gives are not as clear to me. Here is the one I am I currently stuck on.

    1-1/2sin(2x)=sin^3(x)+cos^3(x)/sin(x)+cos(x)

    I'm not sure where to start. Perhaps the right side?

    I hope I formatted it right. I don't know how to make it look pretty.

    Thanks for any help!
    Is this the identity you're trying to prove:

    1-\frac{1}{2sin(2x)}=\frac{sin^3(x)+cos^3(x)}{sin(x)  +cos(x)}

    ??
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  3. #3
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    How do you make it look like that? I'm guessing its code?

    That's very close except the left side is...

    1-(1/2sin(2x))

    Hopefully the parentheses makes it clear.
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  4. #4
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    The first thing you can do to the right side is cancel out the denominator by factoring the numerator as the sum of two cubes.
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  5. #5
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    Quote Originally Posted by repete View Post
    How do you make it look like that? I'm guessing its code?

    That's very close except the left side is...

    1-(1/2sin(2x))

    Hopefully the parentheses makes it clear.
    View this thread:

    http://www.mathhelpforum.com/math-he...-tutorial.html

    There is a PDF file that you can view.
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  6. #6
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    The right side can be simplified down to this:

    \frac{sin^3(x)+cos^3(x)}{sin(x)+cos(x)}=\frac{[sin(x)+cos(x)][cos^2(x)-sin(x)cos(x)+sin^2(x)]}{sin(x)+cos(x)}

    =cos^2(x)-sin(x)cos(x)+sin^2(x)

    =1-sin(x)cos(x)

    I guess your left side must be:

    1-\frac{1}{2}sin(2x)

    =1-\frac{1}{2}(2sin(x)cos(x))

    =1-sin(x)cos(x)
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  7. #7
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    Thanks adkinsjr! I didn't even think about factoring it. I was able to prove it myself with that step in. I'm sure I will be back with more identities. Thanks again!
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