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Math Help - trig ratios

  1. #1
    Senior Member furor celtica's Avatar
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    trig ratios

    is it required to memorize trig ratios such as sin pi/3 = sqrt3 / 2?
    will such ratios be given in examinations?
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  2. #2
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    I'd doubt any exams will require you to memorise these ratios, what I can say is just make sure you are able to re-produce them in exam conditions rather than memorising them individually
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  3. #3
    Senior Member furor celtica's Avatar
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    are you sure of this, and what exactly is 'reproduce them' supposed to mean? i can get up to a certain point but one has to know the exact value where calculator results are not appropriate.
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  4. #4
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    Hello, furor celtica!

    The bad news: Yes, we're expected to know these values.

    The good news: They're not that hard to learn.



    For 45, draw a square (side 1), draw a diagonal.
    Code:
          * - - - - - - *
          |           * |
          |     _   *   |
          |    √2 *     | 1
          |     *       |
          |   *         |
          | * 45       |
          * - - - - - - *
                 1
    Using Pythagorus, the diagonal is \sqrt{2}.

    And we have: . \begin{Bmatrix}\sin45^o &=& \dfrac{opp}{hyp} &=& \dfrac{1}{\sqrt{2}} \\ \\[-4mm]<br />
\cos45^o &=& \dfrac{adj}{hyp} &=& \dfrac{1}{\sqrt{2}} \\  \\[-4mm]<br />
\tan45^o &=& \dfrac{opp}{adj} &=& 1 \end{Bmatrix}




    For 30 and 60, draw an equilateral triangle (side 2) and an alitutude.
    Code:
                  *
                 /|\
                / | \
               /  |  \
            2 /30| _ \ 2
             /    |√3  \
            /     |     \
           / 60  |      \
          * - - - * - - - *
              1       1
    Using Pythagorus, the altitude is \sqrt{3}.


    Then: . \begin{Bmatrix}\sin30^o &=& \dfrac{opp}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \cos30^o &=& \dfrac{adj}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \tan30^o &=& \dfrac{opp}{adj} &=& \dfrac{1}{\sqrt{3}} \end{Bmatrix} . . . \begin{Bmatrix}\sin60^o &=& \dfrac{opp}{hyp} &=& \dfrac{\sqrt{3}}{2} \\ \\[-4mm] \cos60^o &=& \dfrac{adj}{hyp} &=& \dfrac{1}{2} \\ \\[-4mm] \tan60^o &=& \dfrac{opp}{adj} &=& \sqrt{3} \end{Bmatrix}

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  5. #5
    Member great_math's Avatar
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    A trick which i used to remember ratios in my earlier classes(when i was new to trigonometry..)

    For sine ratio

    First, write 0 30 45 60 90 which are the basic ratios.

    Write from 0 to 4 under each degree

    Code:
    0  30   45   60  90
    
    0   1    2    3   4
    Divide each by 4

    Code:
    0  30   45   60  90
    
    0   1    2    3   4    
    -- --   --   --  --
    4   4    4    4   4
    Take the square roots for each..

    Code:
    0  30   45   60  90
    
    0   1    2    3   4    
    -- --   --   --  --
    4   4    4    4   4 
    
    0   1    1   
    
    
    
    
    \sqrt3   1
        --   --   --  --
        2   
    
    
    
    
    \sqrt2    2   1
    We have the sine ratio.. For co-sine ratio take sin 90 as cos 0 and you are done..
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