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Math Help - 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:
    <br />
\sin(30-x) = \frac {3} {\sqrt 2}<br />

    You know that the value of the sin function can vary between -1 and + 1 inclusive, yet if you evaluate  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 \sin\theta = \frac{\sqrt{3}}{2}
    then \theta = \frac{2\pi}{3} , \frac{\pi}{3}

    so in this case \theta = 30 - x if assuming that 30 means 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 \sin(A+B)=\sin{A}\cos{B}+\cos{A}\sin{B}

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

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

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

    =\cos{\left(\frac{\pi}{2} -A\right)}\cos{B}<br />
+\sin{\left(\frac{\pi}{2} -A\right)}\sin{B}

    =\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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