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    how to sketch sin angle +icos angle on an argand diagram?
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    (Original post by crystalong)
    how to sketch sin angle +icos angle on an argand diagram?
    Mind giving us an example? Just evaluate \sin \alpha - that's your real part and then \cos \alpha that's your imaginary part - from which you should be able to plot it on an argand diagram in the usual way.

    For example: \sin \frac{\pi}{6} + i \cos \frac{\pi}{6} = \frac{1}{2} + i \frac{\sqrt{3}}{2} which you should be able to plot by placing a dot 1/2 a unit along the real axis and rt(3)/2 units up the imaginary axis.
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    (Original post by Kvothe the arcane)
    Do you mean cos(angle)+isin(angle)?

    Assuming there is no typo, you should know that  sin (angle) +icos (angle) \equiv i(cos(angle)-isin(angle))

    Evaluate cos(angle)-isin(angle) and multiply the complex number by i and you'll be able to plot that in your argand diagram.
    This doesn't make sense, why would he/she do this? Look at my answer.
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    (Original post by Zacken)
    This doesn't make sense, why would he/she do this? Look at my answer.
    I've deleted my answer. It would have produced the same thing however there were unnecessary steps.
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    (Original post by Kvothe the arcane)
    I've deleted my answer. It would have produced the same thing however there were unnecessary steps.
    It would have, but it's akin to saying "plot a+ib on an argand diagram, first write this as i(-ai + b), evluate b-ai then multiply it by i then plot it v/s just plot a+ib directly".
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    (Original post by Zacken)
    It would have, but it's akin to saying "plot a+ib on an argand diagram, first write this as i(-ai + b), evluate b-ai then multiply it by i then plot it v/s just plot a+ib directly".
    I understand . OP, follow Zacken's help above. It'll lead you to the correct answer.
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    (Original post by crystalong)
    how to sketch sin angle +icos angle on an argand diagram?
    The point lies on a unit circle, inclined at the given angle to the positive x axis
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    (Original post by ODES_PDES)
    The point lies on a unit circle, inclined at the given angle to the positive x axis
    Read the question again. It's not \cos x + i \sin x.
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    I mean how to sketch sin theta +icos theta on an argand diagram?
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    (Original post by Zacken)
    Mind giving us an example? Just evaluate \sin \alpha - that's your real part and then \cos \alpha that's your imaginary part - from which you should be able to plot it on an argand diagram in the usual way.

    For example: \sin \frac{\pi}{6} + i \cos \frac{\pi}{6} = \frac{1}{2} + i \frac{\sqrt{3}}{2} which you should be able to plot by placing a dot 1/2 a unit along the real axis and rt(3)/2 units up the imaginary axis.
    (Original post by crystalong)
    I mean how to sketch sin theta +icos theta on an argand diagram?
    Did you bother reading anything?
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    (Original post by Zacken)
    Did you bother reading anything?
    what if the question states that 0<theta<pie/2 instead of pie/6?
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    (Original post by crystalong)
    what if the question states that 0<theta<pie/2 instead of pie/6?
    How about you show us the question?
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    The question is
    Sketch sin theta +icos theta on an argand diagram, where 0<theta<pie/2. find also its modulus and argument.
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    (Original post by crystalong)
    The question is
    Sketch sin theta +icos theta on an argand diagram, where 0<theta<pie/2. find also its modulus and argument.
    he means do one of these;

    Name:  table.png
Views: 37
Size:  1.6 KB

    also i would suggest drawing it for  \displaystyle  0 &lt; \theta &lt; 2\pi then rubbing out 3/4s of it

    If you do FP2 then you could do like this;


     \displaystyle \text{Let z = complex number (usual notation)}


     \displaystyle   z = \sin\theta + i\cos\theta


     \displaystyle  |z| = |\sin\theta + i\cos\theta|


     \displaystyle  |z| = \sqrt{(\sin\theta)^2 + (\cos\theta)^2}


     \displaystyle  |z| = \sqrt{\sin^2\theta + \cos^2\theta} = \sqrt{1} = 1


     \displaystyle  \therefore |z| = 1

    now its an easy sketch
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    (Original post by DylanJ42)
    he means do one of these;

    Name:  table.png
Views: 37
Size:  1.6 KB

    also i would suggest drawing it for  \displaystyle  0 &lt; \theta &lt; 2\pi then rubbing out 3/4s of it

    If you do FP2 then you could do like this;


     \displaystyle \text{Let z = complex number (usual notation)}


     \displaystyle   z = \sin\theta + i\cos\theta


     \displaystyle  |z| = |\sin\theta + i\cos\theta|


     \displaystyle  |z| = \sqrt{(\sin\theta)^2 + (\cos\theta)^2}


     \displaystyle  |z| = \sqrt{\sin^2\theta + \cos^2\theta} = \sqrt{1} = 1


     \displaystyle  \therefore |z| = 1

    now its an easy sketch
    I got the modulus part. How do you find the argument?
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    (Original post by crystalong)
    I got the modulus part. How do you find the argument?
    Yea, that part I have no idea about

    If I was doing it in an exam I would write 0<x<pi/2 because that was the restriction but I'm really not sure
 
 
 
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