Maths C3 - Trigonometry... Help??

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    (Original post by IrrationalRoot)
    I'm assuming your 'head hurts' (lol) because you always thought \sin had an inverse. If so, answer me this question: if \sin x=0, what is x? Is your answer x=0? Wrong, x=2\pi. In fact x could be any even multiple of \pi.
    I thought that if \sin x=0 then x can equal either  -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0

    (Original post by IrrationalRoot)
    Hopefully you now see why \sin does not have an inverse. If you're given the value of \sin x, you cannot determine x. \sin is not one-to-one.
    No, I know that sin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain  -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x) gets "inverted" to create the function arcsin
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    (Original post by Philip-flop)
    I thought that if \sin x=0 then x can equal either  -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0

    No, I know that sin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain  -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x) gets "inverted" to create the function arcsin
    That is all correct.
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    Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down
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    (Original post by Philip-flop)
    Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down
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    Yes that is correct.

    Alternatively just write:

    \sin(x)=k \Rightarrow x=\arcsin(k)=\alpha
    and
    x=\pi-\alpha
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    (Original post by RDKGames)
    But 0 is a multiply of \pi so it's not wrong, it's just a principal solution, and it doesn't have to be an even multiple. It would be a mess to lead OP to general solutions of sine at this point.
    Idk why I said even, sorry (but this was obviously a mistake...).
    And no, what I'm saying is that I could take the number x=2\pi and give you the information \sin x=0, in which case x=0 is wrong.. This was the point.
    It certainly wouldn't be 'a mess to lead OP to general solutions'. I'm not even teaching them general solutions. I was giving an example to illustrate my point, which it did.
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    (Original post by Philip-flop)
    I thought that if \sin x=0 then x can equal either  -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0



    No, I know that sin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain  -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x) gets "inverted" to create the function arcsin
    Then it doesn't make sense why you don't understand \arcsin(\sin x) \not\equiv x. This is a direct consequence of what you've just said.
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    (Original post by IrrationalRoot)
    Idk why I said even, sorry (but this was obviously a mistake...).
    And no, what I'm saying is that I could take the number x=2\pi and give you the information \sin x=0, in which case x=0 is wrong.. This was the point.
    It certainly wouldn't be 'a mess to lead OP to general solutions'. I'm not even teaching them general solutions. I was giving an example to illustrate my point, which it did.
    Ah okay. I do think it would be a mess based on his understanding, and decided to mention it as your point touches upon them in principle with multiples.
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    1) (cosec2o)^2-(cot2o)^2

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    (Original post by RDKGames)
    Ah okay. I do think it would be a mess based on his understanding, and decided to mention it as your point touches upon them in principle with multiples.
    I think OP can understand multiples... Anyone who has any knowledge of trig functions will understand \sin 2k\pi=0, so this wouldn't confuse OP. It serves as a good example of the fact that \sin is many-to-one. If I just say it's 'many-to-one' with no explanation then that would certainly be more confusing and would likely lead to 'a mess' as you put it.
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    (Original post by Philip-flop)
    I thought that if \sin x=0 then x can equal either  -2\pi , -\pi, 0, \pi, 2\pi and so on. as the graph crosses the x-axis at these points when y=0



    No, I know that sin doesn't have an exact inverse as for it to be a function it must be a one-to-one. Hence why only the domain  -\frac{\pi}{2} \leq x \leq \frac{ \pi}{2} of sin(x) gets "inverted" to create the function arcsin
    I must also point out that this: "for it to be a function it must be a one-to-one"
    is simply wrong. There seems to still be a misunderstanding here. Unless you're talking about the inverse when you say 'it' in which case ok, but see my previous reply.
    Also not sure why so much use of emboldening...
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    (Original post by Philip-flop)
    Are my workings right for question 5? I understand how to get the answer but wasn't too sure how to write everything down
    Name:  C3 - EXE 6E.png
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    Attachment 582432582434
    I know this is completely off topic, but I love your pi. Could you do a tutorial?
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    (Original post by IrrationalRoot)
    Then it doesn't make sense why you don't understand \arcsin(\sin x) \not\equiv x. This is a direct consequence of what you've just said.
    I'm just confused by the equation. I'm sure it's right but I can't seem to understand the layout of it. Sorry

    (Original post by IrrationalRoot)
    I think OP can understand multiples... Anyone who has any knowledge of trig functions will understand \sin 2k\pi=0, so this wouldn't confuse OP. It serves as a good example of the fact that \sin is many-to-one. If I just say it's 'many-to-one' with no explanation then that would certainly be more confusing and would likely lead to 'a mess' as you put it.
    My knowledge of trig functions is, lets just say limited. I have no idea what you mean by the part where you say \sin 2k\pi=0... Where has that K come from??

    I know that  sin2 \pi = 0 though

    (Original post by IrrationalRoot)
    I must also point out that this: "for it to be a function it must be a one-to-one"
    is simply wrong. There seems to still be a misunderstanding here. Unless you're talking about the inverse when you say 'it' in which case ok, but see my previous reply.
    Also not sure why so much use of emboldening...
    How come you only quoted half of my sentence?

    Let me edit a section of what I said so it is a little clearer...
    sin doesn't have an exact inverse as for it (arcsin) to be a function it must be a one-to-one.
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    (Original post by asinghj)
    I know this is completely off topic, but I love your pi. Could you do a tutorial?
    Ah schuckkkks

    Plenty more pi to go around
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    LOOL
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     \sin 2k\pi = 0 just means for every even multiple of  \pi sin will equal 0

    i hate latex
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    (Original post by Philip-flop)
    I'm just confused by the equation. I'm sure it's right but I can't seem to understand the layout of it. Sorry


    My knowledge of trig functions is, lets just say limited. I have no idea what you mean by the part where you say \sin 2k\pi=0... Where has that K come from??

    I know that  sin2 \pi = 0 though


    How come you only quoted half of my sentence?

    Let me edit a section of what I said so it is a little clearer...
    sin doesn't have an exact inverse as for it (arcsin) to be a function it must be a one-to-one.
    k just stands for any integer. So by \sin 2k\pi=0 all I'm saying is that \sin0=\sin2\pi=\sin (-2\pi)= \sin 4\pi=\sin (-4\pi)=...=0 since all these angles are full turns of a circle and will thus have the same sines.

    I mentioned "Unless you're talking about the inverse when you say 'it' in which case ok." I did acknowledge what you said.
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    (Original post by IrrationalRoot)
    k just stands for any integer. So by \sin 2k\pi=0 all I'm saying is that \sin0=\sin2\pi=\sin (-2\pi)= \sin 4\pi=\sin (-4\pi)=...=0 since all these angles are full turns of a circle and will thus have the same sines.

    I mentioned "Unless you're talking about the inverse when you say 'it' in which case ok." I did acknowledge what you said.
    Oh right I see. Yeah I get you now. Learn something everyday

    Thanks for being patient with me btw. I really appreciate it!!
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    (Original post by Philip-flop)
    Oh right I see. Yeah I get you now. Learn something everyday

    Thanks for being patient with me btw. I really appreciate it!!
    Good to hear . Tbh I should be thanking you for being patient with me lol, the number of times today I've had to edit things I've said due to mistakes etc. is ridiculous...
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    (Original post by IrrationalRoot)
    Good to hear . Tbh I should be thanking you for being patient with me lol, the number of times today I've had to edit things I've said due to mistakes etc. is ridiculous...
    Yes it is ridiculous. Especially for someone like you, after having done so brilliantly in STEP. You should be ashamed of yourself
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    So I've just started Chapter 7 of the Edexcel C3 Modular Maths textbook and have already encountered a problem

    It's about proving the 'addition formula' for cos(A-B) = cosAcosB + sinAsinB
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    Ok so I have the following questions to ask about this example...
    1. Why are the coordinates of P and Q... P(cosA, sinA) and Q(cosB, sinB)?
    2. Why is the radius of the circle 1?
    3. Why is the angle POQ equal to (A-B)?? Surely it would be angle POQ = (B-A), is it because using the addition formulae cos(a-b) = cos(b-a)??
    4. How would one know to compare the lengths of PQ^2 from the results of using the cosine rule and the distance between two coordinates for this type of question?
 
 
 
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