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How do you find the exact values of sin(u/2), cos(u/2), and tan(u/2) of tan(u)=-5/8

  • trig -

    since the tangent is negative, angle u is either in quadrant II or IV.
    I made a diagram of a right-angled triangle with a hypotenuse of √89

    we know
    cos 2A = 2cos^2 A -1 or 1 - 2sin^2 A

    so cos u = 2cos^2 (u/2) - 1
    -8/√89 + 1 = 2cos^2 (u/2)
    (√89 - 8)/(2√89) = cos^2 (u/2)
    cos (u/2) = ± √(√89 - 8)/(2√89)

    for the sine:
    cos u = 1 - 2sin^2 (u/2)
    2sin^2 (u/2) = 1 + 8/√89
    sin^2 (u/2) = (√89+8)/(2√89)
    sin (u/2) = ±√(√89+8)/(2√89)

    Now sorting out the ±

    IF u was in quadrant II, then u/2 is in quadrant I and both the sine and cosine would be positive.

    If u was in quadrant IV, the u/2 is in quadrant II
    and the sine would be positive and the cosine negative.

    Since tanx = sinx/cosx
    for tan (u/2) divide the above results, using the appropriate ± values.
    Notice since both the sin (u/2) and cos (u/2) have 2√89 in the denominator, that part will cancel when you do the division.

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