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pre-calculus

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use a sum or difference formula to find the exact of the trigonometric function
cos(-13pi/12)

  • pre-calculus -

    cos(-13pi/12)
    = cos(-13pi/12 + 24pi/12)
    = cos(11pi/12)
    = -cos(pi - 11pi/12)
    = -cos(pi/12)

    now, cos(x/2) = √((1+cos(x))/2)

    -cos(pi/12) = -√((1+cos(pi/6))/2)
    = -√((1+√3/2)/2)
    = -√(2 + √3/4)
    = -1/2 √(2+√3)

    or
    -1/2√2 (1+√3)

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