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Math Help - Trig Double Angle Identities

  1. #1
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    Trig Double Angle Identities

    Stuck on this and don’t have a clue where I am going wrong

    It is given that 3\cos\theta - 2\sin\theta = R\cos(\theta + \alpha), \ where\ R>0\ and\ 0^{\circ} < \alpha < 90^{\circ}

    A) Find the value of R

    B) Show that  \alpha \approx 33.7^{\circ}

    C) Hence write down the maximum value of 3\cos\theta - 2\sin\theta and find a positive value of  \theta at which the maximum value occurs

    NOT AN ACTUAL IDENTITY
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  2. #2
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    Hello, Beard!

    Stuck on this and donít have a clue where I am going wrong.
    It's hard to see your work from here.
    But it looks like you played the {\color{blue}K\spadesuit} instead of the {\color{blue}7\heartsuit}.
    It is given that: . 3\cos\theta - 2\sin\theta \:=\: R\cos(\theta + \alpha),\text{ where }R>0 \text{ and }0^o < \alpha < 90^u

    A) Find the value of R.
    Multiply the left side by \frac{\sqrt{13}}{\sqrt{13}}

    . . \sqrt{13}\left(\frac{3}{\sqrt{13}}\cos\theta - \frac{2}{\sqrt{13}}\sin\theta\right) \;=\;R\cos(\theta + \alpha)


    Let \alpha be an angle such that: . \begin{array}{c}\cos\alpha \:=\:\frac{3}{\sqrt{13}} \\ \\[-4mm] \sin\alpha \:=\:\frac{2}{\sqrt{13}} \end{array}

    Then we have: . \sqrt{13}\underbrace{(\cos\theta \cos\alpha - \sin\theta\sin\alpha)} \:=\:R\cos(\theta + \alpha)

    . . . . . . . . . . . . . . . \sqrt{13}\,\overbrace{\cos(\theta + \alpha)} \:=\:R\cos(\theta + \alpha)

    Therefore: . \boxed{R \:=\:\sqrt{13}}




    Show that: . \alpha \,\approx\, 33.7^o

    Since \sin\alpha \,=\,\tfrac{2}{\sqrt{13}}, then: . \alpha \:=\:\sin^{-1}\left(\tfrac{2}{\sqrt{13}}\right) \quad\Rightarrow\quad \boxed{\alpha\;\approx\; 33.7^o}




    Hence, find the maximum value of 3\cos\theta - 2\sin\theta
    and find a positive value of \theta at which the maximum value occurs.

    The maximum of: . 3\cos\theta - 2\sin\theta \:=\:\sqrt{13}\,\cos(\theta + \alpha) occurs when \cos(\theta + \alpha) \:=\:1

    . . Then the maximum value is \boxed{\sqrt{13}}


    If \cos(\theta + \alpha) \:=\:1, then: . \theta + \alpha \:=\:360^o

    Therefore: . \theta \:=\:360^o - \alpha \:=\:360^o - 33.7^o \quad\Rightarrow\quad\boxed{\theta\;=\;326.3^o}

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