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Math Help - Trigonometric Equations Using Compound Angle Formulae

  1. #1
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    Trigonometric Equations Using Compound Angle Formulae

    Hi all,

    Can someone help with the following question please?

    Solve 5 sin (θ Π/6) = 8 cos θ for values 0 < θ < 2Π

    To begin (using the compound angle formulae), I have:

    5 sin (θ Π/6) = 5 [sinθ cos Π/6 cosθ sin Π/6]


    The trouble is that when I expand the expression I start to get complicated decimal numbers, whereas I wish to maintain Π in the answers, which I want to give in radians not degrees.

    Can anyone help please, or offer an explanatory website that will assist me.

    Thanks.
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  2. #2
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    Quote Originally Posted by Lee66 View Post
    Hi all,

    Can someone help with the following question please?

    Solve 5 sin (θ – Π/6) = 8 cos θ for values 0 < θ < 2Π

    To begin (using the compound angle formulae), I have:

    5 sin (θ – Π/6) = 5 [sinθ cos Π/6 – cosθ sin Π/6]


    The trouble is that when I expand the expression I start to get complicated decimal numbers, whereas I wish to maintain Π in the answers, which I want to give in radians not degrees.

    Can anyone help please, or offer an explanatory website that will assist me.

    Thanks.
    \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}

    \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}

    So, you will obtain:

     5\sin \theta \cos \left(\frac{\pi}{6}\right) - 5\cos\theta\sin\left(\frac{\pi}{6}\right)=8\cos \theta
    \frac{5\sqrt{3}}{2}\sin \theta -\frac52 \cos \theta = 8\cos \theta
    \frac{5\sqrt{3}}{2}\sin \theta=\frac{21}{2} \cos \theta

    Will you be able to continue?
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  3. #3
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    5\sin \left(\theta - \frac{\pi}{6} \right) = 8\cos{\theta}

    5 \left(\sin{\theta} \cos{\frac{\pi}{6}} - \cos{\theta}\sin{\frac{\pi}{6}}\right) = 8\cos{\theta}

    5 \left(\sin{\theta} \cdot \frac{\sqrt{3}}{2} - \cos{\theta} \cdot \frac{1}{2}\right) = 8\cos{\theta}

    \frac{5\sqrt{3}}{2} \sin{\theta} - \frac{5}{2} \cos{\theta} = 8\cos{\theta}

    5\sqrt{3} \sin{\theta} =  21 \cos{\theta}

    75 \sin^2{\theta} = 441 \cos^2{\theta}

    75 \sin^2{\theta} = 441 (1 - \sin^2{\theta})

    \sin^2{\theta} = \frac{441}{516}

    \sin{\theta} = \pm \frac{21}{2\sqrt{129}}

    \theta = \arcsin\left(\frac{21}{2\sqrt{129}}\right) ... solution checks

    \theta = \pi - \arcsin\left(\frac{21}{2\sqrt{129}}\right) ... solution is extraneous

    \theta = \arcsin\left(-\frac{21}{2\sqrt{129}}\right) ... solution is extraneous

    \theta = \pi + \arcsin\left(\frac{21}{2\sqrt{129}}\right) ... solution checks
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  4. #4
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    Quote Originally Posted by skeeter View Post
    ...

    5\sqrt{3} \sin{\theta} =  21 \cos{\theta}

    75 \sin^2{\theta} = 441 \cos^2{\theta}

    ...
    5\sqrt{3}\sin \theta=21 \cos \theta

    You could have divided both sides by \cos \theta to obtain:

    5\sqrt{3}\tan \theta=21

    \tan \theta = \frac{21}{5\sqrt{3}}

    Then, solved for \theta.
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  5. #5
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    could have ... but, it's not always wise to divide by a variable quantity when solving an equation. call it force of habit.
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  6. #6
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    Thanks both for the replies.

    I worked through the problem and also obtained:

     \tan \theta = \frac{21}{5\sqrt{3}}

    How do I solve this for  \theta with an answer containing  \pi ?
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  7. #7
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    Quote Originally Posted by Lee66 View Post
    Thanks both for the replies.

    I worked through the problem and also obtained:

     \tan \theta = \frac{21}{5\sqrt{3}}

    How do I solve this for  \theta with an answer containing  \pi ?

    The range they have provided is 0 < \theta < 2\pi

    So, you can provide the answer in decimal places as \tan ^{-1} \left(\frac{21}{5\sqrt{3}}\right) will not give a standard \pi answers.

    So, for the range provided, the \theta values to 3sf would be:

    \theta = 1.18, 4.32
    ^ This is: \theta = 1.18, (1.18+\pi)

    An accurate answer would be:
    \mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right), \pi + \mathrm{arctan} \left(\frac{21}{5\sqrt{3}}\right)
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