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Calculus

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Find the value of tan2x, (pi/2)<x<pi, given secx=-5/4.

So cosx=-4/5? I've no clue what to do next.

  • Calculus - ,

    According to the given domain, the angle must be in quadrant II,
    We are given cosx = - 4/5.
    remember that cosine is adjacent/hypotenuse, so construct a right angles triangle with base of 4 and hypotenuse of 5, you should recognize the 3-4-5 rightangled triangle so the height (or opposite) is 3 and
    sinx = 3/5
    remember that tangent = sine/cosine
    and tanx = (3/5)/(-4/5) = - 3/4

    have you come across the identiy
    tan 2x = 2tanx/(1 - tan^2 x) ?
    tan2x = 2(-3/4)/(1 - 9/16)
    = (-3/2)/(7/16)
    = -24/7

  • Calculus - ,

    Thanks!

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