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Math Help - Another exact value problem

  1. #1
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    Another exact value problem

    how would you go about solving this:

    if sin x = -4/5, and 270 degrees < x < 360 degrees, then find tan2x
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  2. #2
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    Quote Originally Posted by Kesha View Post
    how would you go about solving this:

    if sin x = -4/5, and 270 degrees < x < 360 degrees, then find tan2x
    \tan (2x) = \frac{2 \tan x}{1 - \tan^2 x}

    From \sin x = - \frac{4}{5} and the fact that x is in the fourth quadrant it follows that \tan x = - \frac{4}{3}.

    Therefore .....
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  3. #3
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    \displaystyle\tan 2x=\frac{2\tan x}{1-\tan^2x} (1)
    If x\in (270^{\circ},360^{\circ})\Rightarrow \cos x>0.
    Then, \cos x=\sqrt{1-\sin^2x}=\frac{3}{5}\Rightarrow\tan x=\frac{\sin x}{\cos x}=-\frac{4}{3}.
    Now, substitute \tan x in (1) and find \tan 2x.
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  4. #4
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    thanks, was easy to do. i just approached it incorrectly.
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