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Thread: 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

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

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

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

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

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

    (a) Find $\displaystyle R$ and $\displaystyle \alpha$

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

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

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

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


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


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




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

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

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  3. #3
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    thank you
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