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Thread: 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
    $\displaystyle \tan (2x) = \frac{2 \tan x}{1 - \tan^2 x}$

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

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