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Math Help - Easy trig problem

  1. #1
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    Easy trig problem

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  2. #2
    Senior Member eumyang's Avatar
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    Maybe it's me, but this didn't seem easy!?!? I'm getting an answer of sin a = 1; is this correct?

    Here's a summary of what I did:
    You have the cosine of a difference identity:
    cos (a-b) = cos\, a\: cos\, b + sin\, a\: sin\, b \quad (Eq1)

    Use the Law of sines to write sin b in terms of sin a:
    sin\, b = \frac{3}{4}sin\, a \quad (Eq2)

    Use the pythagorean identity to write cosine in terms of sine:
    cos\, a = \sqrt{1-sin^2\, a} (Eq3a)
    cos\, b = \sqrt{1-sin^2\, b} (Eq3b)

    Plug (2), (3a), and (3b) into (1), plus the fact that cos (a-b) = 3/4:
    \frac{3}{4} = \sqrt{1-sin^2\, a}\: \sqrt{1-\left(\frac{3}{4}sin\, a\right)^2} + sin\, a\: \left(\frac{3}{4}sin\, a\right) (Eq4)

    Solved for sin a and I got sin a = 1.

    I'm sure I did something wrong here.
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  3. #3
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    In general , if  |ac| = B ~,~ |bc| = A ~,~ \cos(a-b) = X

    The method to find the value of  \sin(a) is as follows :


    First determine which of the angles  a,b is larger by comparing their opposite sides , let's say  a > b .

    Let  d be the other point on line  ab such that  |ac| = |cd| = B

    Then we have


     \angle bcd = a-b , by using cosine formula ,

     |bd| = \sqrt{A^2 + B^2 - 2ABX} ,

    Therefore ,  \sin(a) : A = \sin(a-b) : |bd|


     \sin(a) =\frac{A \sin(a-b)}{ \sqrt{A^2 + B^2 - 2ABX} }
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