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Math Help - Find Exact Value of SinB

  1. #1
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    Post Find Exact Value of SinB

    Heya!
    Ok, here is the exercise:





    Thanks in Advance!
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  2. #2
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    Re: Find Exact Value of SinB

    \cos{\beta} = -\cos{\alpha}

    why?
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    \cos{\beta} = -\cos{\alpha}

    why?
    Where did u get that?
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    Re: Find Exact Value of SinB

    Quote Originally Posted by JohnTerry View Post
    Where did u get that?
    what's the relationship between angles \alpha and \beta ?
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    what's the relationship between angles \alpha and \beta ?
    But... Isn't smth like:
    cosB = π-cosA [in rad]
    cosB = 180-cosA [in degrees]


    How is cosB = -cosA ?
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    Re: Find Exact Value of SinB

    Quote Originally Posted by JohnTerry View Post
    But... Isn't smth like:
    \cos(\beta) = π-\cos(\alpha) [in rad]
    \cos(\beta) = 180-\cos(\alpha) [in degrees]


    How is \cos(\beta)= -\cos(\alpha)?
    \beta = \pi - \alpha

    \cos{\beta} = \cos(\pi - \alpha)

    \cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}

    \cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    \beta = \pi - \alpha

    \cos{\beta} = \cos(\pi - \alpha)

    \cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}

    \cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}
    Then,

    cosA= 3/√45

    cosB= -(3/√45)

    ?
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    Re: Find Exact Value of SinB

    Quote Originally Posted by JohnTerry View Post
    Then,

    cosA= 3/√45

    cosB= -(3/√45)

    ?
    yes ...

    you're welcome.
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    yes ...

    you're welcome.
    Thanks, but in the book the answer is -(√5/5)

    and -(3/√45) is -(3√45/45)
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    Re: Find Exact Value of SinB

    Quote Originally Posted by JohnTerry View Post
    Thanks, but in the book the answer is -(√5/5)

    and -(3/√45) is -(3√45/45)
    -\frac{3}{\sqrt{45}} = -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    -\frac{3}{\sqrt{45}} = -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}
    I'm sorry, but can you explain please how you passed from  -\frac{3}{\sqrt{9} \cdot \sqrt{5}} to -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} ?
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  12. #12
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    Re: Find Exact Value of SinB

    Quote Originally Posted by JohnTerry View Post
    I'm sorry, but can you explain please how you passed from  -\frac{3}{\sqrt{9} \cdot \sqrt{5}} to -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} ?
    -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{3}{3 \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}}

    now rationalize the denominator ...

    -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    -\frac{3}{\sqrt{9} \cdot \sqrt{5}} = -\frac{3}{3 \cdot \sqrt{5}} = -\frac{1}{\sqrt{5}}

    now rationalize the denominator ...

    -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}
    Oh! Thanks!

    But can you please explain one more thing...

    How did u get \cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha} from \cos{\beta} = \cos(\pi - \alpha)

    and \cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}
    from \cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}
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  14. #14
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    Re: Find Exact Value of SinB

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  15. #15
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    Re: Find Exact Value of SinB

    Quote Originally Posted by skeeter View Post
    Thanks again, but how \cos{\beta} = (-1)\cos{\alpha} + (0)\sin{\alpha} = -\cos{\alpha}
    = \cos{\beta} = \cos{\pi}\cos{\alpha} + \sin{\pi}\sin{\alpha}


    What does (-1) and (0) mean?
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