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Math Help - Trigonometry Help

  1. #1
    Junior Member
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    Trigonometry Help

    I have done part of this question but need help finishing it off

    also first time i am using the math code so excuse if i go wrong

    I={4\sin\theta}-{3\cos\theta}
    I={R\sin(\theta-\alpha)}

    Part (a) was to find R and \alpha for which i got
    R = 5
    \alpha = 36.9

    Part (b) Hence solve the equation {4\sin\theta}-{3\cos\theta}=3 to find values of \theta between 0 and 360

    Part (c) write down greatest value for I
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  2. #2
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    Hello, Stylis10!

    I\:=\:{4\sin\theta}-{3\cos\theta}
    I\:=\:R\sin(\theta-\alpha)

    (a) Find R and \alpha

    I got: . R \:= \:5,\;\;\alpha \:\approx \:36.9^o . . . . . Good!
    (b) Hence solve the equation: . 4\sin\theta-3\cos\theta \:=\:3 \;\;\text{ for }0^o \leq \theta < 360^o
    Divide by 5: . \frac{4}{5}\sin\theta - \frac{3}{5}\cos\theta \:=\:\frac{3}{5}

    Let \tan\alpha \:=\:\frac{3}{4}\quad\hdots\quad\text{Then: }\:\cos\alpha \:=\:\frac{4}{5},\;\;\sin\alpha \:=\:\frac{3}{5}\:=\:0.6

    Then we have: . \cos\alpha\sin\theta - \sin\alpha\cos\theta \:=\:\frac{3}{5}\quad\Rightarrow\quad \sin(\theta -\alpha) \:=\:0.6

    . . \theta - \alpha \;=\;\sin^{-1}(0.6) \quad\Rightarrow\quad\theta \;=\;\sin^{-1}(0.6) + \alpha


    Since \alpha \:=\:\sin^{-1}(0.6), we have: . \theta \;=\;2\sin^{-1}(0.6)


    Since \sin^{-1}(0,6) \:\approx\:36.9^o,\;143.1^o,\;\;\theta \;=\;73.8^o,\;286.2^o




    (c) Find the greatest value for I.
    In part (a), we had: . I \;=\;5\sin(\theta-36.9^o)

    The maximum value is: . I \:=\:5

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  3. #3
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    thank you
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