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Math Help - help with expressions

  1. #1
    Junior Member
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    help with expressions

    values of two of the three expressions cos (x), sin (x), and tan (x), has not been set. Decide these values.

    sin(x) = -((2)/(5^0.5)) , 3n/2<x<2n
    cos(x)^2 + sin(x)^2 = 1

    cos(x) = (1-sin(x)^2)^0.5 = (1-(-2/(5^0.5))^2)^0.5 = (9/5)^0.5

    tan(x) = ((-2/(5^0.5))/((9/5)^0.5) = (-10^0.5)/5


    When I leave it to my teacher I get error, does anyone know what the problem is? He says that you should show 3n/2<x<2n in sin, cos and tan in any way.


    By n I mean pi
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  2. #2
    Super Member

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    Hello, paulaa!

    Your arithmetic is off . . .


    Values of two of the three expressions \cos\theta,\:\sin\theta,\:\tan\theta have not been set.
    Decide these values.

    . . \sin\theta \,=\,\text{-}\dfrac{2}{\sqrt{5}},\;\;\frac{3\pi}{2} < x < 2\pi

    \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \cos\theta \:=\:\pm\sqrt{1-\sin^2\!\theta}

    \cos\theta \:=\:\pm\sqrt{1 - \left(\text{-}\frac{2}{\sqrt{5}}\right)^2} \:=\:\pm\sqrt{1 - \frac{4}{5}} \:=\:\pm\sqrt{\frac{1}{5}} \;=\;\pm\dfrac{1}{\sqrt{5}}


    Since \,\theta is in Quadrant 4, \cos\theta is positive: . \boxed{\cos\theta \:=\:\frac{1}{\sqrt{5}}}

    Then: . \tan\theta \:=\:\dfrac{\sin\theta}{\cos\theta} \:=\:\dfrac{\text{-}\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} \quad\Rightarrow\quad \boxed{\tan\theta \:=\:-2}

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