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Math Help - way to express ans

  1. #1
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    way to express ans

    Can anyone help to check if my communication skill for part B is good enough please?

    The question is as following:

    Part a:
    transform  2 \cos(\theta) + \sin(\theta) into the form r \sin (\theta + \alpha), where r > 0 and 0 \deg < \alpha < 90 \deg .

    >>>>> my ans obtained is
    [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

    Part B
    Given that - \alpha \<= \theta \ <= 180\deg - \alpha, find the value of k such that the equation

    2\cos \theta + 3 \sin \theta = k
    has a unique solution.



    >>>>>>>>>>my answer:

    first I sketched (not draw) a graph of [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

    then :
    Given - \alpha \<= \theta \ <= 180\deg - \alpha
    ie - 33.7 \deg <= \theta \ <= 146.3 \deg
    From the sketched graph, it is only possible for k to have a unique solution, where - \alpha \<= \theta \ <= 180\deg - \alpha when [tex]\sqrt{13} [/Math]
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  2. #2
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    Quote Originally Posted by ling_c_0202
    Can anyone help to check if my communication skill for part B is good enough please?

    The question is as following:

    Part a:
    transform  2 \cos(\theta) + \sin(\theta) into the form r \sin (\theta + \alpha), where r > 0 and 0 \deg < \alpha < 90 \deg .

    >>>>> my ans obtained is
    [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]
    We want to rewrite

    <br />
 2 \cos(\theta) + \sin(\theta)=r \sin (\theta + \alpha)<br />

    So we may use the trig identity for the sin of the sum of two angles to
    expand the RHS:

    <br />
 2 \cos(\theta) + \sin(\theta)=r (\sin (\theta) \cos( \alpha)+\cos(\theta) \sin(\alpha))<br />

    so we require:

    <br />
2=r \sin(\alpha)<br />
    <br />
1=r \cos(\alpha)<br />

    Squaring and adding these equations give:

    <br />
2^2+1^2=r^2 ((\cos (\alpha))^2 + (\sin(\alpha))^2)=r^2<br />

    so r=\sqrt{5}.

    Also dividing the first equation by the second gives:

    <br />
2=\tan(\alpha)<br />

    so \alpha = \arctan(2) \approx 1.107 radian, or \approx 63.4 ^{\circ}

    RonL
    Last edited by CaptainBlack; April 15th 2006 at 12:57 PM.
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  3. #3
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    Quote Originally Posted by ling_c_0202
    Can anyone help to check if my communication skill for part B is good enough please?

    The question is as following:

    Part a:
    transform  2 \cos(\theta) + \sin(\theta) into the form r \sin (\theta + \alpha), where r > 0 and 0 \deg < \alpha < 90 \deg .

    >>>>> my ans obtained is
    [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]
    Of course your solution corresponds to rewriting  2 \cos(\theta) +3 \sin(\theta) in the form  r \sin (\theta + \alpha).

    RonL
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  4. #4
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    Quote Originally Posted by ling_c_0202
    [Part B
    Given that - \alpha \<= \theta \ <= 180\deg - \alpha, find the value of k such that the equation

    2\cos \theta + 3 \sin \theta = k
    has a unique solution.
    We are interested in what values of k result in a unique solution
    to the equation:

    2\cos \theta + 3 \sin \theta=\sqrt{13} \sin(\theta+\alpha) = k

    for - \alpha <= \theta  <= 180\deg - \alpha.

    Put \phi=\theta+\alpha, then the question becomes: For which
    values of k does the equation:

    <br />
\sqrt{13} \sin(\phi) = k<br />

    have a unique solution for 0 <= \phi  <= 180\deg.

    The solution is:

    <br />
\phi=\arcsin(k/\sqrt{13})<br />
,

    which is unique for \phi in the specified range when:

    <br />
k/ \sqrt{13}= 1<br />
,

    or:

    <br />
k= \sqrt{13}<br />
,

    all other values of k give two solutions in the range, this is illustrated in the
    attached plot.

    RonL
    Attached Thumbnails Attached Thumbnails way to express ans-trig-plot1.jpg  
    Last edited by CaptainBlack; April 15th 2006 at 11:34 PM.
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  5. #5
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    Arrow

    Quote Originally Posted by ling_c_0202
    Can anyone help to check if my communication skill for part B is good enough please?

    The question is as following:

    Part a:
    transform  2 \cos(\theta) + \sin(\theta) into the form r \sin (\theta + \alpha), where r > 0 and 0 \deg < \alpha < 90 \deg .

    >>>>> my ans obtained is
    [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

    Part B
    Given that - \alpha \<= \theta \ <= 180\deg - \alpha, find the value of k such that the equation

    2\cos \theta + 3 \sin \theta = k
    has a unique solution.



    >>>>>>>>>>my answer:

    first I sketched (not draw) a graph of [tex]\sqrt{13} \sin (\theta + 33.7 deg)[/Math]

    then :
    Given - \alpha \<= \theta \ <= 180\deg - \alpha
    ie - 33.7 \deg <= \theta \ <= 146.3 \deg
    From the sketched graph, it is only possible for k to have a unique solution, where - \alpha \<= \theta \ <= 180\deg - \alpha when [tex]\sqrt{13} [/Math]
    If you know how to graph or sketch sine and cosine curves then here is one way to do your two problems here.

    1.) Express [2cos(theta) +sin(theta)] into [r*sin(theta +Alpha)], where Alpha is a constant and it is between 0 and 90deg, and r is greater than zero.

    You graph y = 2cos(theta) and y = sin(theta) on the same (y,theta) cartesian axes.
    The sum of the two y's is the y of r*sin(theta +Alpha).

    Alpha, by the way, is the horizontal shift. Let's call it angle A.

    In those two graphs, you see that
    ---at theta=0, r*sin(theta +A) = 2.
    ---at theta=90deg, r*sin(theta +A) = 1
    ---at theta=180deg, r*sin(theta +A) = -2.
    ---Etc.

    So we use two of those.

    When theta=0,
    r*sin(0+A) = 2
    sin(A) = 2/r -----(1)

    When theta=90deg,
    r*sin(90+A) = 1
    r*[sin(90deg)cosA +cos(90deg)sin(A) = 1
    r*[1*cosA +0*sinA] = 1
    cosA = 1/r -----(2)

    (1) divided by (2),
    sinA/cosA = (2/r)/(1/r)
    tanA = 2
    A = arctan(2) = 63.435 degrees ----***
    Substitute that into, say, (2),
    cos(63.435deg) = 1/r
    r = 1/cos(63.435deg) = 2.236 -----***

    Therefore, 2cos(theta) +sin(theta) = 2.236sin(theta +63.435), in degrees. ----answer.

    ----------------------
    2.) Find k so that [2cos(theta) +3sin(theta) = k] has unique solution.
    where theta is between -Alpha and (180 -Alpha)deg.

    You found
    2cos(theta) +3sin(theta) = sqrt(13)*sin(theta +33.7deg)
    Which is correct.

    So Alpha = 33.7deg'
    In the graph of y = sqrt(13)*sin(theta +33.7deg), based from the graphs of y=2cos(theta) and y=3sin(theta), the horizontal shift is to the left of the origin (0,0).
    Meaning,
    ---the "start"--where it is zero--of the resulting sine curve is at theta = -33.7 deg.
    ---so the "end"--where is goes down to zero again--of the curve that is above the theta-axis (horizontal axis) is at (180 -33.7) deg.

    The positive half of the graph of a sine curve is in the first half of the first cycle or period.
    One cycle of y = sqrt(13)*sin(theta +33.7deg) is 360 degrees long, so the first half of that is from 0 to 180deg.

    Meaning ,further, that in the interval theta = [-33.7,(180-33.7)], the curve of [2cos(theta) +3sin(theta)] is above the theta- or horizontal axis.

    As seen from y = sqrt(13)*sin(theta +33.7deg), the amplitude is sqrt(13). So the curve went from 0 at theta=(-33.7deg), went to a maximum of sqrt(13) as it goes to the right, then came down to zero again at theta=(180 -33.7)deg.

    Therefore, for 2cos(theta) +3sin(theta) = k, k=[0,+sqrt(13)], or k must be from zero to positive sqrt(13), so that there will be unique solutions to the said equation in the interval theta = [-33.7deg,(180-33.7)deg]. ---------answer.
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  6. #6
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    Quote Originally Posted by ticbol


    Therefore, for 2cos(theta) +3sin(theta) = k, k=[0,+sqrt(13)], or k must be from zero to positive sqrt(13), so that there will be unique solutions to the said equation in the interval theta = [-33.7deg,(180-33.7)deg]. ---------answer.
    That any value of k \in [0,\sqrt{13}] does not give a unique solution to:

    <br />
2 \cos(\theta) +3\sin (\theta) = k<br />
,

    is illustrated by the attached plot. This shows that k=1 gives two
    solutions in the range, this is generally the case except when k=\sqrt{13},
    when the two solutions merge to give a unique solution.

    RonL
    Attached Thumbnails Attached Thumbnails way to express ans-trig-plot.jpg  
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  7. #7
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    Quote Originally Posted by CaptainBlack
    That any value of k \in [0,\sqrt{13}] does not give a unique solution to:

    <br />
2 \cos(\theta) +3\sin (\theta) = k<br />
,

    is illustrated by the attached plot. This shows that k=1 gives two
    solutions in the range, this is generally the case except when k=\sqrt{13},
    when the two solutions merge to give a unique solution.

    RonL
    Umm, yes, you are right. When k=1, there are two values of theta in that positive first half of the graph of y = 2cos(theta) +3sin(theta).
    I missed that.

    But k is not +,-sqrt(13) as you said before in your not-yet-edited answer then.

    Oh, well, who does not make mistake.
    I do. I do make mistakes always. Sometimes I could correct them before someone else points them.

    That's why this is a forum.
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  8. #8
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    Quote Originally Posted by ticbol
    Umm, yes, you are right. When k=1, there are two values of theta in that positive first half of the graph of y = 2cos(theta) +3sin(theta).
    I missed that.

    But k is not +,-sqrt(13) as you said before in your not-yet-edited answer then.

    Oh, well, who does not make mistake.
    I do. I do make mistakes always. Sometimes I could correct them before someone else points them.

    That's why this is a forum.
    I had misread the range for the solution, and was looking a full cycle of
    the sine curve - just shows a picture always helps.

    RonL
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  9. #9
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    thank you

    Thank you very much for your kindly help.

    I have got it!! THZZ
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