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Thread: Mixed Problems

  1. #1
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    Mixed Problems

    Hey i have some mixed problems i cant get my head around, i was wondering would any of you know the possible solutions to the following:

    sin(30-X)=3÷root 2...finding the values between 0 and 360 degrees.
    and
    Prove sin(A+B)=sinAcosB+cosAsinB

    Thank you guys!
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  2. #2
    MHF Contributor ebaines's Avatar
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    For the first question you have:
    $\displaystyle
    \sin(30-x) = \frac {3} {\sqrt 2}
    $

    You know that the value of the sin function can vary between -1 and + 1 inclusive, yet if you evaluate $\displaystyle 3/\sqrt 2$ what do you get?

    As for the proof of the sum of angles formula, see: Proof of the sum and difference formulas.
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  3. #3
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    Hey thanks buddy, is there a shorter way of prooving that identity?
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  4. #4
    Super Member bigwave's Avatar
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    recall that when $\displaystyle \sin\theta = \frac{\sqrt{3}}{2}$
    then $\displaystyle \theta = \frac{2\pi}{3} $, $\displaystyle \frac{\pi}{3}$

    so in this case $\displaystyle \theta = 30 - x$ if assuming that $\displaystyle 30 $ means $\displaystyle 30^o $ then you can calculate the values for x
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  5. #5
    Super Member bigwave's Avatar
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    to prove $\displaystyle \sin(A+B)=\sin{A}\cos{B}+\cos{A}\sin{B}$

    one way to do this to use the identity
    $\displaystyle
    \sin{\theta} = \cos{\left(\frac{\pi}{2} - \theta\right)}
    $then

    $\displaystyle \sin{\left(A+B\right)} = \cos{\left(\frac{\pi}{2} -(A+B)\right)} $

    $\displaystyle = \cos{\left(\left(\frac{\pi}{2} -A\right)-B\right)}$

    $\displaystyle =\cos{\left(\frac{\pi}{2} -A\right)}\cos{B}
    +\sin{\left(\frac{\pi}{2} -A\right)}\sin{B}$

    $\displaystyle =\sin{A}\cos{B}+\cos{A}\sin{B}$
    Last edited by bigwave; May 24th 2010 at 11:04 AM. Reason: more steps added
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