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Math Help - need trig help

  1. #1
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    need trig help

    10) Given that tan = -1/7 and sin > 0, use the x-y-r method to find sin and cos.

    11) Sketch a 240 angle in standard position. Label a point on its terminal side, draw the right triangle in use, indicate the direction of opening, and give all six trig function values for 240.

    12) Sketch a 90 angle in standard position. Label a point on its terminal side and use the x-y-r method to compute the six trig function values for 90.

    13) In what quadrant(s) does terminate if sin > 0 and cos < 0?
    Q1 Q2 Q3 Q4

    14) In what quadrant(s) does terminate if tan < 0?
    Q1 Q2 Q3 Q4

    15) In what quadrant(s) does terminate if cos < 0?
    Q1 Q2 Q3 Q4

    17) Use a fundamental identity to find cos if sin = 1/5, that terminates in Q2. Show work.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, gabby!

    Here's some help . . .


    10) Given that \tan\theta = -\frac{1}{7} and \sin\theta > 0,
    use the x-y-r method to find: . \sin\theta and \cos\theta

    \tan\theta is negative in Q2 and Q4.
    \sin\theta is positive in Q1 and Q2.
    . . Hence, \theta is in Q2.

    We have: . \tan\theta \:=\:-\frac{1}{7} \:=\:\frac{y}{x}

    Since \theta is in Q2: . x = -7,\;y = 1
    . . Then: . r \:=\:\sqrt{x^2+y^2} \:=\:\sqrt{(-7)^2 + 1^2} \:=\:\sqrt{50} \:=\:5\sqrt{2}


    Therefore: . \begin{Bmatrix} \sin\theta &=&\dfrac{y}{r} &=&\dfrac{1}{5\sqrt{2}} &=& \dfrac{\sqrt{2}}{10} \\ \\[-4mm]<br />
\cos\theta &=& \dfrac{x}{r} &=& \dfrac{-7}{5\sqrt{2}} &=& - \dfrac{7\sqrt{2}}{10} \end{Bmatrix}




    11) Sketch a 240 angle in standard position.
    Label a point on its terminal side, draw the right triangle in use,
    indicate the direction of opening, and give all six trig function values for 240.
    Code:
                    |
                    |
                -1  |
          - - + - - * - - - -
            _ |    /|
          -√3 |   / |
              |  /2 |
              | /   |
              |/    |
              *     |

    We have: . x = -1,\;y = -\sqrt{3},\;r = 2

    Then: . \begin{Bmatrix} \sin\theta &=& \frac{y}{r} &=&-\frac{\sqrt{3}}{2} \\ \\[-4mm] \cos\theta &=& \frac{x}{r} &=& -\frac{1}{2} \\ \\[-4mm] \tan\theta &=& \frac{y}{x} &=& \sqrt{3} \\ \vdots & & \vdots & & \vdots  \end{Bmatrix}




    13) In what quadrant(s) does \theta terminate if \sin\theta > 0 and \cos\theta < 0?

    \sin\theta is positive in Q1 and Q2.
    \cos\theta is negative in Q2 and Q3.

    Therefore, \theta terminates in Q2.




    17) Use a fundamental identity to find \cos\theta
    if \sin\theta = \frac{1}{5}, and \theta terminates in Q2.

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

    We have: . \cos\theta \;=\;\pm\sqrt{1 - \left(\frac{1}{5}\right)^2} \;=\;\pm\sqrt{\frac{24}{25}} \;=\;\pm\frac{2\sqrt{6}}{5}


    Since \theta is in Q2, \cos\theta is negative.

    Therefore: . \cos\theta \;=\;-\frac{2\sqrt{6}}{5}

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