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Can someone explain to me how to use the quadrant diagram for c3 trig? watch

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    ...please. (It's the All Silly Tom Cats in each quadrant thing.)

    Basically i have the c3/c4 green books, but didnt have the c1/c2 books so im not aware on how to use this method for trig, as for c2 trig i would just draw a graph..but drawing a sec and cosec graph is too tiring, so i guess they make us use this. Umm a problem example will be needed probably, so the one here is:

    sec\theta = -2.5
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    If sec x = -2.5, then cos x = -0.4. But what this doesn't (necessarily) mean is that x = arccos(-0.4) - you need to check the boundaries on theta. Basically, if you draw one of these ASTC diagrams, it tells you which quadrant each trig function is positive or negative in. As cos x < 0, you know that it's going to be in either the "S" or the "T" quadrant (right?). Work out arccos(0.4), which will correspond to the first quadrant, and draw yourself a little diagram and use symmetry to work out the answers.

    This make much sense?
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    The fact that i don't know how you get cos x = -4, and what arccos is etc might not be helping, as these stuff are further on in the book, lemme give it all a read over and it might be better.
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    (Original post by T_Bag)
    The fact that i don't know how you get cos x = -4, and what arccos is etc might not be helping, as these stuff are further on in the book, lemme give it all a read over and it might be better.
    arccos is the same as cos-1, the button on your calculator meaning "inverse cos". If sec x = -2.5, then cos x = 1/-2.5 = -0.4.
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    I think i know what your talking about. Basically the quadrant diagram represents in which quadrant each function is positive. For example, 0-90, the first quadrant, s c t are all positive. 90-180, s. 180-270, t. 270-360 c.

    The way i use it, is simply to calculate the other angles that satisfy the equation... in each quadrant, starting from 0-90..

    theta, then 180-theta, 180+theta, and finally 360-theta.

    Now, say you want to calculate cosx=0.2..

    cos-1 0.2=78.5

    Now we look to the c quadrant, 360-78.5=281.5

    The only thing that changes, is if cos=-0.2, in which case you would tick not theta and cos, but the sin and tan quadrant.

    Cos-1 0.2=78.5.... so 180-78.5=101.5 and 180+78.5=258.5

    Reading back over this, it'll probably confuse you more than it will help you :P But hey, it works for me
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    sec(theta)=-2.5..... Firstly, 1/-2.5=-0.4 Now, since it is -ve, we'll take theta of Cosx=0.4, and just reverse which quadrants we use.... so now instead of theta and 360-t being the angles, we want 180+t and 180-t.

    Cos-1 0.4=66.4 180+/-66.4=246 and 114

    Which is exactly what gb said, without the arccos :P Hope it makes anyway
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    Aye i kinda get it now a wee bit, but why dont you do cos^-1( -0.4)=\theta

    Why is the negative sign of 0.4 ignored?

    Ahh and cheers with the arcoss thing gen! excess names to remember :rolleyes:
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    It's a C3 thing, really; but arccos can only give out values between 0 and pi. So if you're looking for anything else (say you were trying to solve for -2pi < x < 2pi), you won't get it. Simplest thing to do is take the inverse cos of the positive value (which means you'll always get a value in the first quadrant).

    I say "simplest"; it's arbitrary. If you always took arccos of the negative value, you'd always get a value in the second quadrant. I just find the first quadrant easier to deal with.

    Edit: and I dislike the notation "arccos" myself, but cos^-1 takes too long to type. :p:
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    (Original post by generalebriety)
    It's a C3 thing, really; but arccos can only give out values between 0 and pi. So if you're looking for anything else (say you were trying to solve for -2pi < x < 2pi), you won't get it. Simplest thing to do is take the inverse cos of the positive value (which means you'll always get a value in the first quadrant).

    I say "simplest"; it's arbitrary. If you always took arccos of the negative value, you'd always get a value in the second quadrant. I just find the first quadrant easier to deal with.

    Edit: and I dislike the notation "arccos" myself, but cos^-1 takes too long to type. :p:
    Hahaha so true...

    Ok, so in C3 i always use +ve values for inverse anything(sin,cos or tan)?
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    (Original post by T_Bag)
    Hahaha so true...

    Ok, so in C3 i always use +ve values for inverse anything(sin,cos or tan)?
    You use whatever you feel comfortable with. I don't use this method at all, I just remember being taught to use positive values and then read off the 'correct' answers from the CAST diagram.
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    One can quickly identify solutions by sketching a graph of f(\theta ) against \theta for the trigonometric function in question, and then identifying where one's value of f(\theta ) corresponds to a value of \theta. (In the original question, we want to find the value of \theta that corresponds to f(\theta ) = -0.4)
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    There is no rule that you NEED to reverse the sign, for example it would be quite correct to go.

    Cosx=-0.8 Arccos-0.8=143. That gives you theta, the top right quadrant, and now we go to the "cos" quadrant, bottom right, for 360-theta, 217. 143 and 217 is the correct answer. However, sometimes (i can't think of when), it is easier to reverse the sign, Arccos0.8=36.9, reverse the quadrants we use, so now top left and bottom left, "sin" and "tan" respectively, 180+36.9=217, 180-36.9=143.

    In essence, all you need to know is for Sin, t and 180-t. Cos, t and 360-t. Tan, t and 180+t. In some circumstance, you find it's easier to calculate arccos of the reverse sign, or are given the actual value in a non calc paper, you use the other two equations.
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    Ahh what i do now is, use my calc to work out the arc's of everything..and plot the relevant graph, see how many points they fall on. With time and practise a sketch will probably not be necessary.
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    (Original post by T_Bag)
    Ahh what i do now is, use my calc to work out the arc's of everything..and plot the relevant graph, see how many points they fall on. With time and practise a sketch will probably not be necessary.
    Well the CAST diagram is really an adaptation of this idea, except you go round in circles rather than along a line. I must admit, I find it pretty necessary for almost all trig that I do. When questions ask for solutions in a range, I think it would be much quicker than your method. But, whatever you feel most comfortable with I guess
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    (Original post by coffeym)
    Well the CAST diagram is really an adaptation of this idea, except you go round in circles rather than along a line. I must admit, I find it pretty necessary for almost all trig that I do. When questions ask for solutions in a range, I think it would be much quicker than your method. But, whatever you feel most comfortable with I guess
    ahh i see, aybe ill try and learn the cast diagram, i am really comfy with this method though
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    (Original post by T_Bag)
    ahh i see, aybe ill try and learn the cast diagram, i am really comfy with this method though
    Well, feel free to carry on using it.

    Purely for reference, this is what I use:

     \sin \theta = a \Rightarrow \theta = n\pi + (-1)^n\sin^{-1} (a)

     \cos \theta = a \Rightarrow \theta = 2n\pi\pm\cos^{-1} (a)

     \tan \theta = a \Rightarrow \theta = n\pi +\tan^{-1} (a)

    ...or arcsin, arccos, arctan, of course.
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    (Original post by generalebriety)
    Well, feel free to carry on using it.

    Purely for reference, this is what I use:

     \sin \theta = a \Rightarrow \theta = n\pi + (-1)^n\sin^{-1} (a)

     \cos \theta = a \Rightarrow \theta = 2n\pi\pm\cos^{-1} (a)

     \tan \theta = a \Rightarrow \theta = n\pi +\tan^{-1} (a)

    ...or arcsin, arccos, arctan, of course.
    damn... tricks of the genious's hey? ...howd you come up with that? derive it yourself? :p: ...i could honestly believe that. Btw, what does the n stand for?
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    (Original post by T_Bag)
    damn... tricks of the genious's hey? ...howd you come up with that? derive it yourself? :p: ...i could honestly believe that. Btw, what does the n stand for?
    I did derive it myself, and I've seen it all over the place since, it's not hard by looking at the graphs. n is any integer. n=0 will give you one solution (or in the case of cos, two solutions), n=1 will give you the next solution/two solutions, then n=2, n=3, ..., and same for n=-1, n=-2...
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    (Original post by generalebriety)
    I did derive it myself, and I've seen it all over the place since, it's not hard by looking at the graphs. n is any integer. n=0 will give you one solution (or in the case of cos, two solutions), n=1 will give you the next solution/two solutions, then n=2, n=3, ..., and same for n=-1, n=-2...
    How long do i keep using different integers for? Until i have enough solutions with the range?
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    (Original post by T_Bag)
    How long do i keep using different integers for? Until i have enough solutions with the range?
    Yep.

    With practice it'll be obvious. If sin x = sin 0.2, and you want to find the possible values for x, 0 < x < 3pi, then obviously you don't go down as far as -pi - 0.2 (where n = -1) or up as far as 4pi + 0.2 (n = 4). :p: And you get the solutions x = 0.2, pi - 0.2, 2pi + 0.2, 3pi - 0.2 (for n = 0, 1, 2, 3).

    Notice that this is easily seen from the graph.
 
 
 
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