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Math Help - Special Triangles Question

  1. #1
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    Special Triangles Question

    I am looking for help with this question from my grade 11 advanced functions course:

    csc θ = -17/15 where 270 ≤ θ ≤ 360 and
    cot β = - 3/4 where 90 ≤ β ≤ 180.

    Find the exact value of sin(θ + β). Show all the work.
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  2. #2
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    Better yet, why don't you show all of YOUR work, i.e. what you have tried and where exactly you are stuck... Then you might get the help you need...
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  3. #3
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    This what I have done:

    cscθ = -17/15 where 270 ≤ θ ≤ 360

    sinθ = 1/cscθ
    sinθ = 1/(-17/15)
    sinθ = 1/1 / (-17/15)
    sinθ = 1/1 * (-15/17)
    sinθ = -15/17
    θ = sin^-1 (-15/17)

    ** I don't know what to do from here because if I solve for θ completely I won't have an EXACT value **


    cot β = - 3/4 where 90 ≤ β ≤ 180

    tanβ = 1/cotβ
    tanβ = 1/(-3/4)
    tanβ = 1/1 / (-3/4)
    tanβ = 1/1 * (-4/3)
    tanβ = -4/3
    β = tan^-1 (-4/3)

    ** I don't know what to do from here because if I solve for β completely I won't have an EXACT value **


    From this point I'm not sure how to find an exact value for sin(θ + β) because I do not have values for θ or β
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  4. #4
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    For starters, you should know that \displaystyle \sin{(\theta \pm \beta)} \equiv \sin{(\theta)}\cos{(\beta)} \pm \cos{(\theta)}\sin{(\beta)}.

    You don't need to solve for \displaystyle \theta or \displaystyle \beta, you only need to solve for \displaystyle \sin{(\theta)}, \cos{(\theta)}, \sin{(\beta)} and \displaystyle \cos{(\beta)}. You will need to make use of the Pythagorean Identity.
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  5. #5
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    I knew about the sum and difference formulas and that I would have to use the sum formula for sin in order to solve the final step, but I'm not sure what you mean by pythagorean identities. I'm not sure how to solve for sin(\theta), cos(\theta), sin(\beta), or cos(\beta).
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  6. #6
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    Quote Originally Posted by starshine84 View Post
    This what I have done:
    cscθ = -17/15 where 270 ≤ θ ≤ 360
    cot β = - 3/4 where 90 ≤ β ≤ 180
    ** I don't know what to do from here because if I solve for β completely I won't have an EXACT value
    From that given you know that \theta\in IV.
    So \sin \left( \theta  \right) = \frac{{ - 15}}{{17}}\;\& \,\cos \left( \theta  \right) = \sqrt {1 - \left( {\frac{15}{17}} \right)^2 }=\frac{8}{{17}}

    Moreover, \beta\in II so \sin \left( \beta  \right) = \frac{4}{5}\;\& \,\cos \left( \beta  \right) = \frac{{ - 3}}{5}.
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  7. #7
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    So would it then be...

    sin(\theta+\beta) = sin\theta cos\beta + cos\theta sin\beta
    sin(\theta+\beta) = (-15/17)(-3/4) + (8/17)(4/5)
    sin(\theta+\beta) = (45/68) + (32/85)
    sin(\theta+\beta) = 225/340 + 128/340
    sin(\theta+\beta) = 353/340

    ??
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  8. #8
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    Looks good to me
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  9. #9
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    That's awesome, thank you so much for all of your help!
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