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    I always create new threads when im stuck on an A level ques. So from now on ill just post here - and people can help. First ques is about CAST circles with radians from C2 - does anyone know about cast circles?
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    Many people on here will know about CAST diagrams, You're going to have to be a bit more specific.
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    (Original post by kiiten)
    I always create new threads when im stuck on an A level ques. So from now on ill just post here - and people can help. First ques is about CAST circles with radians from C2 - does anyone know about cast circles?
    Watching this is very helpful, I find. :yep:
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    (Original post by B_9710)
    Many people on here will know about CAST diagrams, You're going to have to be a bit more specific.
    Wait ill post the question and my working
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    (Original post by kiiten)
    Wait ill post the question and my working
    Yes please.
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    Question 2.k)
    So i started from pi/2 to get 1/6pi as a solution. Is this wrong? NOTE: i drew the angle from 0 and when i start from 0 i get the right answer? (Sorry if that doesnt make sense)

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    (Original post by kiiten)
    Question 2.k)
    So i started from pi/2 to get 1/6pi as a solution. Is this wrong? NOTE: i drew the angle from 0 and when i start from 0 i get the right answer? (Sorry if that doesnt make sense)

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    Remember your CAST diagram rules? You should be doing \cos u = -\frac{1}{2} but then your "key-angle" is \arccos \frac{1}{2} = \frac{\pi}{3} and not \arccos -\frac{1}{2}. Then because -\frac{1}{2} is negative, your solutions will be in the second and third quadrant. So \pi + \frac{\pi}{3} and...? You tell me.
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    (Original post by Zacken)
    Remember your CAST diagram rules? You should be doing \cos u = -\frac{1}{2} but then your "key-angle" is \arccos \frac{1}{2} = \frac{\pi}{3} and not \arccos -\frac{1}{2}. Then because -\frac{1}{2} is negative, your solutions will be in the second and third quadrant. So \pi + \frac{\pi}{3} and...? You tell me.
    I'm confused

    cos (x + pi/2) = -1/2
    let u = x + pi/2

    so cosu = -1/2
    u = 2/3 pi

    is that right so far? - I don't understand what you mean after you mention "key-angle"
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    (Original post by kiiten)
    so cosu = -1/2
    u = 2/3 pi
    This bit is not quite correct. It should be:

    So u = pi + cos^(-1) (1/2) and u = pi - cos^(-1) (1/2).

    Remember when you're doing CAST diagrams, you take the inverse trig of modulus of the RHS.
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    [QUOTE=Zacken;63278041]This bit is not quite correct. It should be:

    So u = pi + cos^(-1) (1/2)

    where did the pi come from and where did the minus from the 1/2 go?
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    (Original post by kiiten)

    where did the pi come from and where did the minus from the 1/2 go?
    Did you watch the video that linked you? That should explain everything.
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    (Original post by Zacken)
    Did you watch the video that linked you? That should explain everything.
    yeah ive watched the video but it only seems to explain the basics. E.g. when I do the inverse of cos (-1/2) = 2/3 pi but you said something about adding pi ?
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    (Original post by kiiten)
    yeah ive watched the video but it only seems to explain the basics. E.g. when I do the inverse of cos (-1/2) = 2/3 pi but you said something about adding pi ?
    Let me run through an example for you. \sin x = -\frac{1}{2}. Now I'm going to use something called my key angle, which is \sin^{-1} \frac{1}{2} = 30^{\circ}.

    Now, I also know that -\frac{1}{2} is negative, so my solutions will be in the second and fourth quadrant. Which means I need to do: 180^{\circ} + 30^{\circ} and 360^{\circ}.

    If it was \sin x = \frac{1}{2}, my key-angle would still be \sin^{-1} \frac{1}{2} = 30^{\circ} but this time \frac{1}{2} is positive, so my solutions will be in the third and first quadrant, so my solutions are 30^{\circ} and 180 - 30^{\circ}.

    You getting the gist of the method?
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    A little bit - i havent learnt this way before. So for the key angle do you always use that sin-1 (1/2) = 30?
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    (Original post by kiiten)
    A little bit - i havent learnt this way before. So for the key angle do you always use that sin-1 (1/2) = 30?
    You should know that sin-1 (1/2) = 30. This is because sin(30) = 1/2.
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    (Original post by Zacken)
    You should know that sin-1 (1/2) = 30. This is because sin(30) = 1/2.
    I asked my friend for help - i think i know where i went wrong. If you look at the diagram i attached earlier i was calculating the solution from the bound pi/2 rather than from 0 so my answer would be slightly less. I.e. the angle started from 0 and stopped after pi/2 so you still use the full angle. If that makes any sense?

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    General cast circle ques

    E.g. if the bounds are from pi/2 to 5/2 pi and the angle is 2/3 pi do you calculate the solution from 0 (green arrow) or pi/2 (red arrow)?

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    (Original post by kiiten)
    General cast circle ques

    E.g. if the bounds are from pi/2 to 5/2 pi and the angle is 2/3 pi do you calculate the solution from 0 (green arrow) or pi/2 (red arrow)?

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    If the bounds are between pi/2 and 5pi/2 then it makes sense to only look for solutions between pi/2 and 5pi/2. So you would look for solutions from the red arrow.

    But angles will still start from 0.

    Your question isn't completely clear so it may help if you post an example question if you're still not sure.
    So the red arrow.
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    (Original post by notnek)
    If the bounds are between pi/2 and 5pi/2 then it makes sense to only look for solutions between pi/2 and 5pi/2. So you would look for solutions from the red arrow.

    But angles will still start from 0.

    Your question isn't completely clear so it may help if you post an example question if you're still not sure.
    So the red arrow.
    If the angle is 2/3 pi and i use the red arrow the answer is wrong - i get 1/6 pi. But if you said the angle starts from 0 shouldnt it be from the green arrow because the angle stops after the bound?

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    (Original post by kiiten)
    If the angle is 2/3 pi and i use the red arrow the answer is wrong - i get 1/6 pi. But if you said the angle starts from 0 shouldnt it be from the green arrow because the angle stops after the bound?

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    yes always measure angles from the green arrow

    what notnek is saying is that solutions between from where the red arrow starts until a full rotation later (ie 1.25 cirlces) are the only valid ones
 
 
 
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