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Math Help - Precal Test Tomorrow

  1. #1
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    Precal Test Tomorrow

    Seems that others have similar problems that I do. I've finished up most of my review packet, just a few things I don't quite get yet. Wondering if someone here could clarify these two questions. Then all I have left is to face the daunting task of saving myself from my own stupidity.

    1. Find the exact value of the expression. Do not use a calculator.

    csc^2 70 degrees - tan^2 20 degrees

    multiple choice

    a) 2 b) 0 c) -1 d) 1

    I think the others I am figuring out right now.
    Last edited by Gerbilkit; January 10th 2008 at 07:21 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gerbilkit View Post
    Seems that others have similar problems that I do. I've finished up most of my review packet, just a few things I don't quite get yet. Wondering if someone here could clarify these two questions. Then all I have left is to face the daunting task of saving myself from my own stupidity.

    1. Use a calculator to solve the equation on the interval 0 <= theta <= 2pi

    csc^2 70 degrees - tan^2 20 degrees

    multiple choice

    a) 2 b) 0 c) -1 d) 1

    I think the others I am figuring out right now.
    you should know that \csc^2 70 = \frac 1{\sin^2 70}, and that you can plug in to your calculator
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  3. #3
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    I need the exact values please. No calculators.
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  4. #4
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    Im confused. Your question states to use a calculator making it a trivial problem but now you say you cannot use a calculator?
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  5. #5
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    Oh erm oops. Starting typing out the wrong question. Sorry about the confusion.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    ok, here's the trick. it might take a while for us to go through this with me giving you hints, so i'll just give you a possible solution.

    things to know:

    (1) \csc x = \frac 1{\sin x}

    (2) \sec x = \frac 1{\cos x}

    (3) \tan^2 x = \sec^2 x - 1


    (4) the sine of an angle is equal to the cosine of its compliment (complimentary angles are angles that sum to 90 degrees).

    so, for instance: \sin 30 = \cos 60, \sin 45 = \cos 45, \sin 10 = \cos 80, etc

    particularly what we will use here is: \cos 20 = \sin 70 \implies \cos^2 20 = \sin^2 70


    solution:

    \csc^2 70 - \tan^2 20 = \frac 1{\sin^2 70} - \left( \sec^2 20 - 1 \right)

    = \frac 1{\sin^2 70} + 1 - \sec^2 20

    = \frac 1{\sin^2 70} + 1 - \frac 1{\cos^2 20}

    = \frac 1{\sin^2 70} + 1 - \frac 1{\sin^2 70}

    = 1

    there are other solutions here. this is just the one that occurred to me first. maybe sixstringartist can come up with another one
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  7. #7
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    Hello, Gerbilkit!

    1. Find the exact value of the expression. Do not use a calculator.

    . . \csc^2\!70^o - \tan^2\!20^o

    (a)\;2\qquad (b)\;0\qquad  (c)\:-1\qquad (d)\;1
    Note the identity: . \sin70^o \:=\:\cos20^o


    We have: . \csc^2\!70^o - \tan^2\!20^o \;\;=\;\;\frac{1}{\sin^2\!70^o} - \frac{\sin^2\!20^o}{\cos^2\!20^o} \;\;=\;\;\frac{1}{\cos^2\!20^o} - \frac{\sin^2\!20^o}{\cos^2\!20^o}

    . . =\;\;\frac{1-\sin^2\!20^o}{\cos^2\!20^o} \;\;=\;\;\frac{\cos^2\!20^o}{\cos^2\!20^o} \;\;=\;\;{\color{blue}\boxed{1}} . . . answer (d)



    Too slow again . . .
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Too slow again . . .
    yes. but the identities you used were nicer. i mean who can remember that 1 + \tan^2 x = \sec^2 x, seriously
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    yes. but the identities you used were nicer. i mean who can remember that 1 + \tan^2 x = \sec^2 x, seriously
    Well I did but I didn't know about the complementary angle thingamajig. That certainly worked out nicely. The only concern I have is that possibly I wasn't supposed to know that, and was supposed to use a different method to solve it, well if that's the case too bad. Thanks again.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gerbilkit View Post
    ... I didn't know about the complementary angle thingamajig... The only concern I have is that possibly I wasn't supposed to know that, and was supposed to use a different method to solve it...
    i doubt you can solve this without a calculator if you didn't know that. but of course, everytime i say something like that, someone comes along and proves me wrong, so i'm waiting ...
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