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Thread: Choose the point on the terminal side of theta.

  1. #1
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    Choose the point on the terminal side of theta.

    Hello! I ran onto these twos problem on one of my assignments. We're only a few weeks into this semester and I'm not sure how to solve these without a calculator.. Could someone please help me start to solve it or solve them for me(step-by-step)? Thank you so much!

    "Without using a calculator, choose the point on the terminal side of theta.
    1). theta = 5pi/4"

    and

    "Evaluate without using a calculator.
    2). sin theta, if cos theta=2/5 and tan theta<0"

    Thank you very much!
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  2. #2
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    Re: Choose the point on the terminal side of theta.

    If you have a point,$(x,y)$ that is $r$ units out along a line that is $\theta$ radians from the x-axis then

    $(x,y) = (r \cos(\theta, r \sin(\theta))$

    For problem (1)

    $r=1,~\theta = \dfrac {5\pi}{4}$

    $(x,y) = \left(1\cdot \cos\left(\dfrac {5\pi}{4}\right),1\cdot \sin\left(\dfrac {5\pi}{4}\right) \right)$

    Can you evaluate that without a calculator? Use the Pythagorean theorem if you need to.

    For problem (2)

    note that $\sin^2(\theta) + \cos^2(\theta)=1$

    and select from the two possible values of $\sin(\theta)$ using the information on the tangent
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  3. #3
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    Re: Choose the point on the terminal side of theta.

    Quote Originally Posted by Venusian View Post
    Hello! I ran onto these twos problem on one of my assignments. We're only a few weeks into this semester and I'm not sure how to solve these without a calculator.. Could someone please help me start to solve it or solve them for me(step-by-step)? Thank you so much!
    "Without using a calculator, choose the point on the terminal side of theta.
    1). theta = 5pi/4"
    Had I written this question, I would have expected you to know that $\dfrac{5\pi}{4}$ is in III and and has reference angle of $\dfrac{\pi}{4}$.
    Therefore any point on its terminal ray has ordinates $\bf{t}\left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2} \right),~\bf{t}<0$

    That is as random as it gets.
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