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Math Help - solve 8 sin (theta)-15cos(theta)=8.5

  1. #1
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    solve 8 sin (theta)-15cos(theta)=8.5

    i am trying to workout part (ii) illustrate your method

    i already have done part (i)

    (i) Express 8 Sin\theta - 15Cos\theta

    in the format R.Sin(\theta + \alpha)

    \theta = tan^{-1} \frac {15}{8}

    \theta = 61.9^0

    \theta=360-61.9 = 298.1^0<br />



    (ii) using your answer to part (i) solve

    8 Sin\theta - 15Cos\theta = 8.5

    for  0^0\leq\theta\leq360^0

    (illustrate your method)

    any help much appreciated
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  2. #2
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    Hello, decoy808!

    i already have done part (i) . . . . but you didn't finish

    (i) Express: . X \:=\:8\sin\theta - 15\cos\theta in the format: R\sin(\theta + \alpha)
    This is how I would do it . . .


    We have: . \sqrt{8^2+15^2} \:=\:\sqrt{289} \:=\:17

    Multiply X by \tfrac{17}{17}\!:\;\;\frac{17}{17}\left(8\sin\thet  a - 15\cos\theta\right) \;=\;17\left(\tfrac{8}{17}\sin\theta - \tfrac{15}{17}\cos\theta\right)

    Let \alpha be in a right triangle with: . opp = 15,\:adj= 8,\:hyp = 17
    . . Then: . \cos\alpha = \tfrac{8}{15},\;\sin\alpha = \tfrac{15}{17}

    Then we have: . X \;=\;17(\cos\alpha\sin\theta - \sin\alpha\cos\theta)

    From a Compound-Angle Identity: . \boxed{X \;=\;17\sin(\theta - \alpha)}


    Since \sin\alpha \:=\: \tfrac{15}{17}\:=\:0.882352941, then: . \alpha \,\approx\,61.9^i




    (ii) Using your answer to part (i), solve: . 8\sin\theta - 15\cos\theta \:=\: 8.5\;\;\text{ for }0^o\leq\theta\leq360^o

    So we have: . 8\sin\theta -15\cos\theta \:=\:8.5

    . . which becomes: . 17\sin(\theta - 69.1^o) \:=\:8.5


    Then: . \sin(\theta - 61.9^o) \:=\:0.5\quad\Rightarrow\quad \theta - 61.9^o \:=\:30^o,\:150^o

    . . Therefore: . \boxed{\theta \;=\;91.9^o,\;211.9^o}

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