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Math Help - Pythagorean identities

  1. #1
    Member purplec16's Avatar
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    Pythagorean identities

    Use Pythagorean identities to write the expression as an integer.

    tan^2 4β - sec^2 4β

    Please help me i dont know what to do...
    Last edited by purplec16; January 26th 2010 at 05:55 PM.
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  2. #2
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    Quote Originally Posted by purplec16 View Post
    Use Pythagorean identities to write the expression as an integer.

    tan^2 4β - sec^2 4β

    Please help me i dont know what to do...
    hint ...

    1 + \tan^2(4\beta) = \sec^2(4\beta)
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  3. #3
    Member purplec16's Avatar
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    lol...omg i'm still lost...
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  4. #4
    Member purplec16's Avatar
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    1+tan^2(4\beta)-sec^2(4\beta)<br />
    tan^2(4\beta)-sec^2(4\beta)=-1<br />

    is this the next step, i dont understand how you get it to equal to an integer...i.e. get rid of the beta
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  5. #5
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    Skeeter has given you the answer. You don't need to get rid of beta.

     1 + \tan^2(4\beta) = \sec^2(4\beta)

     1 + \tan^2(4\beta) - \sec^2(4\beta)=0

     \tan^2(4\beta) - \sec^2(4\beta)=-1
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  6. #6
    Member purplec16's Avatar
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    Oh ok...wow i knew how to do it then...
    so what if it was something like
    4 tan^2(\beta)-4sec^2(\beta)
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  7. #7
    Member purplec16's Avatar
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    Oh ok...wow i knew how to do it then...
    so what if it was something like
    4 tan^2 (\beta)-4 sec^2 (\beta)
    what would happen in that case?
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  8. #8
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    Quote Originally Posted by purplec16 View Post
    Use Pythagorean identities to write the expression as an integer.
    \tan^2(4\beta) - \sec^2(4\beta)=-1

    -1 is an integer so you are done.

    For the next question

    4 tan^2 (\beta)-4 sec^2 (\beta)

    4( tan^2 (\beta)- sec^2 (\beta))

    4( -1)
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  9. #9
    Member purplec16's Avatar
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    Sorry to bother you but would u be able to assist me in solving something like this:

    \frac{cot^2\alpha-4}{cot^2\alpha-cot\alpha-6}
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  10. #10
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    What are you trying to do? Just simplify?

    \frac{\cot^2\alpha-4}{\cot^2\alpha-\cot\alpha-6}

    make x =\cot\alpha

    \frac{x^2-4}{x^2-x-6}

    \frac{(x-2)(x+2)}{(x-3)(x-2)}

    \frac{x+2}{x-3}

    \frac{\cot\alpha+ 2}{\cot\alpha -3}
    Last edited by pickslides; January 26th 2010 at 06:47 PM. Reason: bad latex
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  11. #11
    Member purplec16's Avatar
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    yes simplifry the expression
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  12. #12
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    Quote Originally Posted by purplec16 View Post
    yes simplifry the expression

    Is done in the previous post.

    Can also say

    \frac{\cot\alpha+ 2}{\cot\alpha -3} =  1+\frac{5}{\cot\alpha -3}
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  13. #13
    Member purplec16's Avatar
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    Quick Question for this expression:

    5sin^2(\theta/4)+5cos^2(\theta/4)

    how do u get rid of the (\theta/4) to make it equal to 5? i understand that it will be equal to 5
    is it that i dont have to worry about the theta and jus simplify it equal to five
    i.e. 5(sin^2(\theta/4)+cos^2(\theta/4))
    5(1)
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  14. #14
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    \sin^2\left(\frac{\theta}{4}\right)+\cos^2\left(\f  rac{\theta}{4}\right) = 1

    Now factor out the 5 and follow what I did in post #8.
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  15. #15
    Member purplec16's Avatar
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    Ok, thank you so much, I did that
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