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Math Help - Some exercises

  1. #1
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    Some exercises

    If someone could help me slove a few exercises ,I realy need them?
    I would need them by Wednesday.
    And sorry me english is bad.
    Here they are:
    http://img337.yfrog.com/img337/2927/mata2b.png

    Thanks!
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  2. #2
    Super Member Deadstar's Avatar
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    Will help with a couple... Please show any working so we know what you have done. I'll give you the answers as well but you'll need to some of your own working...

    First question.
    (1-\sin(\alpha)-\cos(\alpha))(1-\sin(\alpha)+\cos(\alpha)) = 1 - 2\sin(\alpha) + \sin^2(\alpha) - \cos^2(\alpha) by expanding.

    Then use 1 = \sin^2(\alpha) + \cos^2(\alpha).

    ANSWER: 2

    Second question.

    Use the fact that a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) and replace a and b with \sin(3\alpha) and \cos(3\alpha).

    ANSWER: 1 + \tan^2(\alpha) (however this can be further simplified, I'll leave that up to you, use the hint in the next question)

    Third question.

    \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha), hence for this question... \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha) = -2 so \sin(\alpha) = -2\cos(\alpha). Substitute in the equation.

    ANSWER: -\frac{1}{3}
    Last edited by Deadstar; March 29th 2010 at 12:54 PM.
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  3. #3
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    Hello, Thaurus!


    (3)\;\;\frac{(1-\sin\alpha - \cos\alpha)(1 - \sin\alpha + \cos\alpha)}{\sin\alpha(\sin\alpha - 1)}
    The numerator is:

    . . 1-\sin\alpha + \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha - \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha

    . . . . =\;\;(1-2\sin\alpha + \sin^2\!\alpha) -\cos^2\!\alpha \;\;=\;\;(1-\sin\alpha)^2 - (1-\sin^2\!\alpha)

    . . . . =\;\;(1-\sin\alpha)^2 - (1-\sin\alpha)(1 + \sin\alpha) \;\;=\;\;(1-\sin\alpha)\bigg[(1-\sin\alpha) - (1 + \sin\alpha)\bigg]

    . . . . =\;\;-2\sin\alpha(1-\sin\alpha)


    The fraction becomes: . \frac{-2\sin\alpha(1-\sin\alpha)}{\sin\alpha(\sin\alpha - 1)} \;\;=\;\;\frac{2\sin\alpha(\sin\alpha-1)}{\sin\alpha(\sin\alpha-1)} \;\;=\;\;2




    (2)\;\;\frac{\sin^4\!3u - \cos^4\!3u}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u

    \text{Factor: }\;\frac{(\sin^2\!3u - \cos^2\!3u)\overbrace{(\sin^2\!3u + \cos^2\!3u)}^{\text{This is 1}}}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u

    . . . . =\;\;1 + \tan^2\!3u \;\;=\;\;\sec^2\!3u


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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Soroban View Post

    \text{Factor: }\;\frac{(\sin^2\!3u - \cos^2\!3u)\overbrace{(\sin^2\!3u + \cos^2\!3u)}^{\text{This is 1}}}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u

    . . . . =\;\;1 + \tan^2\!3u \;\;=\;\;\sec^2\!3u


    Is it not just \tan^2{\alpha}? (or u as you have it)
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Soroban View Post

    . . 1-\sin\alpha + \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha - \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha

    . . . . =\;\;(1-2\sin\alpha + \sin^2\!\alpha) -\cos^2\!\alpha \;\;=\;\;(1-\sin\alpha)^2 - (1-\sin^2\!\alpha)

    . . . . =\;\;(1-\sin\alpha)^2 - (1-\sin\alpha)(1 + \sin\alpha) \;\;=\;\;(1-\sin\alpha)\bigg[(1-\sin\alpha) - (1 + \sin\alpha)\bigg]
    Just a thought...

    A much simpler way (i think) to compute the above is...

    1-\sin\alpha +  \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha -  \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha

    . . . . =\;\;1-2\sin\alpha +  \sin^2\!\alpha -\cos^2\!\alpha \;\;

    = \sin^2(\alpha) + \cos^2(\alpha) - 2\sin(\alpha) + \sin^2(\alpha) - \cos^2(\alpha)

     = -2\sin(\alpha) + 2\sin^2(\alpha)

     = -2\sin(\alpha)(1 - \sin(\alpha))
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  6. #6
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    Hello again, Thaurus!

    1690. Evaluate: . \frac{\sin\alpha + \cos\alpha}{\cos\alpha - \sin\alpha}\,\:\text{ where }\:\tan\alpha\,=\,-2

    We have: . \frac{\sin\alpha + \cos\alpha}{\cos\alpha - \sin\alpha}

    Divide top and bottom by \cos\alpha\!:\quad \frac{\dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\cos\alpha}{\cos\alpha}}{\dfrac{\cos\alpha}  {\cos\alpha} - \dfrac{\sin\alpha}{\cos\alpha}} \;=\; \frac{\tan\alpha + 1}{1 - \tan\alpha}


    Substitute \tan\alpha = \text{-}2\!:\;\;\frac{(\text{-}2)+1}{1-(\text{-}2)} \;=\;-\frac{1}{3}




    2.\;\;\frac{\tan135^o}{\tan225^o}\cdot\sin300^o - \cos^2(\text{-}300^o)

    We have: . \frac{-1}{+1}\left(-\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)^2\;\;=\;\;\frac{\sqrt{3}}  {2} - \frac{1}{4}

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  7. #7
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    Thanks!
    I did it myself , am I am right?
    Attached Thumbnails Attached Thumbnails Some exercises-math.bmp  
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  8. #8
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Thaurus View Post
    Thanks!
    I did it myself , am I am right?
    Yep thats right!

    If it was an assessment question you might want to show a few more steps of working (in my opinion!). Something like...
    1 + \cot^2(x) = 1+ \frac{1}{\tan^2(x)} = 1 + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)}
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