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11 November, 16:21

How do you evaluate sin (13π/12) ?

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  1. 11 November, 19:32
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    13pi/12 lies between pi and 2pi, which means sin (13pi/12) < 0

    Recall the double angle identity,

    sin^2 (x) = (1 - cos (2x)) / 2

    If we let x = 13pi/12, then

    sin (13pi/12) = - sqrt[ (1 - cos (13pi/6)) / 2]

    where we took the negative square root because we expect a negative value.

    Now, because cosine has a period of 2pi, we have

    cos (13pi/6) = cos (2pi + pi/6) = cos (pi/6) = sqrt[3]/2

    Then

    sin (13pi/12) = - sqrt[ (1 - sqrt[3]/2) / 2]

    sin (13pi/12) = - sqrt[2 - sqrt[3]]/2
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