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Math Help - Trigonometry problem , easy for someone who knows math :)

  1. #1
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    Trigonometry problem , easy for someone who knows math :)

    Imageshack - 87978225.jpg thats the problem please help i have a test tomorrow and if i fail i will loose one year please help
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  2. #2
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    Is this problem for studying for the test? Or is it a take-home test?
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  3. #3
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    Answer

    Alpha=x.. k?
    cos x=5/13
    x is between 270,360 so x is negative of the form (360-x).

    cos x= cos -x
    therefore, we can take cos x as acute.
    but sin x is negative
    sin360-x = - sin x

    sin(60-x)=sin60*cosX -sin(-X)cos60
    =root3 *5/13 + 12/13*1/2
    =5root3 + 12
    ------------
    26


    cos(45+x)=cos45cosX-sin45sin(-X)
    =1/root2*5/13 + 1/root2*12/13
    = 17
    ------
    13root2
    Last edited by umangarora; August 20th 2010 at 12:15 PM.
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  4. #4
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    Your addition of angles formula is the right way to go about this. The restriction on x just tells you the sign of sin(x), where you use Pythagorean theorem to find sin(x) without actually needing to solve for x. I would double-check your trig there, especially the sin(60).
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  5. #5
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    Hello, TheGame!

    \cos\theta \:=\:\frac{5}{13},\;\;\theta \in Q4

    Find: . (a)\;\sin(60^o-\theta)
    . . . . . (b)\;\cos(\theta + 45^o)

    \cos\theta \:=\:\dfrac{5}{13} \:=\:\dfrac{adj}{hyp}

     \theta is in a right triangle with: adj = 5,\;hyp = 13

    Pythagorus tells us that: opp = \pm 12

    Since  \theta is in Q4,\;\sin\theta \:=\:\text{-}\frac{12}{13}



    (a)\;\sin(60^o - \theta) \;=\;\sin60^o\cos\theta - \cos60^o\sin\theta

    . . . . . . . . . . . . =\;\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{5}  {13}\right) - \left(\dfrac{1}{2}\right)\left(\text{-}\dfrac{12}{13}\right)

    . . . . . . . . . . . . =\;\dfrac{5\sqrt{3}+12}{26}



    (b)\;\cos(\theta + 45^o) \;=\;\cos\theta\cos45^o - \sin\theta\sin45^o

    . . . . . . . . . . . . =\;\left(\dfrac{5}{13}\right)\left(\dfrac{1}{\sqrt  {2}}\right) - \left(\text{-}\dfrac{12}{13}\right)\left(\dfrac{1}{\sqrt{2}}\ri  ght)

    . . . . . . . . . . . . =\;\dfrac{17}{13\sqrt{2}}
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  6. #6
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    Also,i use sin^2 x + cos^2 x=1.. another form of pythagoras.. but straight to the point..
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