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Math Help - Exact value of sin 17pi/12

  1. #1
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    Exact value of sin 17pi/12

    12.a) Determine an exact value for sin 17pi/12.

    b) Determine the answer to part a) using a different method.

    So as of now I only know one way to do it.. which is using the compound angle formula. This is what I did:

    sin(x+y) = sinxcosy+cosxsiny
    sin(2pi/3 + 3pi/4) = sin(2pi/3)cos(3pi/4)+cos(2pi/3)sin(3pi/4)
    =(sqr3/2)(-1/sqr2) + (-1/2)(1/sqr2)
    =-sqr3/2sqr2 + -1/2sqr2
    =-sqr3-1/2sqr2

    This is the correct answer. Any ideas on what a second way of finding the exact value of sin 17pi/12 would be? The easier the better haha.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kmjt View Post
    12.a) Determine an exact value for sin 17pi/12.

    b) Determine the answer to part a) using a different method.

    So as of now I only know one way to do it.. which is using the compound angle formula. This is what I did:

    sin(x+y) = sinxcosy+cosxsiny
    sin(2pi/3 + 3pi/4) = sin(2pi/3)cos(3pi/4)+cos(2pi/3)sin(3pi/4)
    =(sqr3/2)(-1/sqr2) + (-1/2)(1/sqr2)
    =-sqr3/2sqr2 + -1/2sqr2
    =-sqr3-1/2sqr2

    This is the correct answer. Any ideas on what a second way of finding the exact value of sin 17pi/12 would be? The easier the better haha.
    You can check the answer with your calculator:

    Code:
    >sin(17*pi/12)
        -0.965926 
    >
    >(sqrt(3)/2)*(-1/sqrt(2))+(-1/2)*(1/sqrt(2))
        -0.965926 
    >
    Your poor use of brackets makes it difficult to tell if you have simplified correctly.

    For a second method use:

    \sin(17 \pi/12)=\sin(\pi +5\pi/12)=-\sin(5\pi/12)

    Now proceed as before.

    CB
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  3. #3
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    Hello, kmjt!

    12 (a) Determine an exact value for: . \sin\tfrac{17\pi}{12}

    . . (b) Determine the answer to part (a) using a different method.

    Use the identity: . \cos2\theta \:=\:1-2\sin^2\!\theta \quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}}


    Since \tfrac{17\pi}{12} is in Quadrant 3, \sin\tfrac{17\pi}{12} is negative.


    We have: . \sin\frac{17\pi}{12} \;=\;-\sqrt{\frac{1-\cos\frac{17\pi}{6}}{2}} \;=\;-\sqrt{\frac{1-\cos\frac{5\pi}{6}}{2}} \;=\;-\sqrt{\frac{1-(\text{-}\frac{\sqrt{3}}{2})}{2}}

    . . . . . . . . =\;-\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}} \;=\;-\sqrt{\frac{2+\sqrt{3}}{4}} \;=\; -\frac{\sqrt{2+\sqrt{3}}}{2}


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  4. #4
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    I'm not quite sure whats going on there How is the stuff changing in the square root? Like at first the top is 1 - cos 17pi6 then in the next part its cos 5pi/6?
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