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Math Help - Trig value question

  1. #1
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    Trig value question

    I would like to know if I am doing these correctly. I am supposed to find exact solutions of pi/6,pi/4 for these problems.
    cos(11pi/12), tan(pi/12), sin(5pi/12) This is one of the ones I did.

    Cos(11pi/12)= Cos(3pi/4 + pi/6) - Sin(3pi/4 + pi/6)
    Cos(3pi/4) Cos(pi/6) - Sin(3pi/4) Sin(pi/6)
    -sqrt(2)/2 *sqrt(3)/2 - Sqrt(2)/2 *1/2 = -sqrt(6)/2 -sqrt(2)/2
    Also if this is correct, do I use cotangent with tan.

    Thank you,
    Keith Stevens
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  2. #2
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    Quote Originally Posted by kcsteven View Post
    I would like to know if I am doing these correctly. I am supposed to find exact solutions of pi/6,pi/4 for these problems.
    cos(11pi/12), tan(pi/12), sin(5pi/12) This is one of the ones I did.

    Cos(11pi/12)= Cos(3pi/4 + pi/6) - Sin(3pi/4 + pi/6)
    Cos(3pi/4) Cos(pi/6) - Sin(3pi/4) Sin(pi/6)
    -sqrt(2)/2 *sqrt(3)/2 - Sqrt(2)/2 *1/2 = -sqrt(6)/2 -sqrt(2)/2
    Also if this is correct, do I use cotangent with tan.

    Thank you,
    Keith Stevens
    I'm curious, where did you get that first line from?

    Your second line is wrong.
    cos \left ( \frac{3\pi}{4} + \frac{\pi}{6} \right ) = cos \left ( \frac{3\pi}{4} \right ) cos \left ( \frac{\pi}{6} \right ) - sin \left ( \frac{3\pi}{4} \right ) sin \left ( \frac{\pi}{6} \right )

    and

    sin \left ( \frac{3\pi}{4} + \frac{\pi}{6} \right ) = sin \left ( \frac{3\pi}{4} \right ) cos \left ( \frac{\pi}{6} \right ) + sin \left ( \frac{\pi}{6} \right ) cos \left ( \frac{3\pi}{4} \right )

    I would use tan \left ( \frac{\pi}{12} \right ) = tan \left ( \frac{1}{2} \cdot \frac{\pi}{6} \right )

    for the tangent problem.

    -Dan
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