Maths C3 - Trigonometry... Help??

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    (Original post by RDKGames)
    As long as you remember \sin^2 \theta + \cos^2 \theta \equiv 1 you can pretty much derive the rest from it so you don't necessarily have to remember all of them.
    Oh really? Can you explain. I'm not sure I see how the other identities are derived from that. I see them all as separate identities which is why I find it difficult to remember them
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    (Original post by Philip-flop)
    Oh really? Can you explain. I'm not sure I see how the other identities are derived from that. I see them all as separate identities which is why I find it difficult to remember them
    Take the original identity:
    \displaystyle \cos^2 x + \sin^2 x \equiv 1

    Of course from here, you can rearrange for sine or cosine, or leave it for 1.

    If you divide everything by cosine squared you get:

    \displaystyle 1+\frac{\sin^2 x}{\cos^2 x} \equiv \frac{1}{\cos^2 x} \Rightarrow 1 + \tan^2 x \equiv \sec^2 x

    Again, you can rearrange for whatever you need.

    Lastly, take the original and divide by sine squared instead. You get:

    \displaystyle \frac{\cos^2 x}{\sin^2 x} + 1 \equiv \frac{1}{\sin^2 x} \Rightarrow \cot^2 x + 1 \equiv \csc^2 x

    Again, you can rearrange for whichever term you need.
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    (Original post by RDKGames)
    Take the original identity:
    \displaystyle \cos^2 x + \sin^2 x \equiv 1

    Of course from here, you can rearrange for sine or cosine, or leave it for 1.

    If you divide everything by cosine squared you get:

    \displaystyle 1+\frac{\sin^2 x}{\cos^2 x} \equiv \frac{1}{\cos^2 x} \Rightarrow 1 + \tan^2 x \equiv \sec^2 x

    Again, you can rearrange for whatever you need.

    Lastly, take the original and divide by sine squared instead. You get:

    \displaystyle \frac{\cos^2 x}{\sin^2 x} + 1 \equiv \frac{1}{\sin^2 x} \Rightarrow \cot^2 x + 1 \equiv \csc^2 x

    Again, you can rearrange for whichever term you need.
    Thank you so much! I probably should have known that considering there was an example on it in the book. But things always seem much clearer after hearing another perspective
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    Ok so I think I've hit another brick wall I have no idea how to even start this question ...

    Q) Simplify the following expression...

     \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta
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    (Original post by Philip-flop)
    Ok so I think I've hit another brick wall I have no idea how to even start this question ...

    Q) Simplify the following expression...

     \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta
    Use the fact that sec^2(x)=1+tan^2(x)
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    (Original post by Philip-flop)
    Ok so I think I've hit another brick wall I have no idea how to even start this question ...

    Q) Simplify the following expression...

     \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta
    Factorise it. Let x=\sec^2 \theta if you have to and then treat tan as some constant number. Then apply some identities to the bracket to see if it simplifies.

    This is like a simple case of a^2-2ab+b^2=(a-b)^2
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    (Original post by Philip-flop)
    Ok so I think I've hit another brick wall I have no idea how to even start this question ...

    Q) Simplify the following expression...

     \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta
    You know that (x-y)^2 = x^2 - 2xy + y^2 in your case, that looks exactly like yours with x = \sec^2 \theta and y = \tan^2 \theta.
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    (Original post by Zacken)
    You know that (x-y)^2 = x^2 - 2xy + y^2 in your case, that looks exactly like yours with x = \sec^2 \theta and y = \tan^2 \theta.
    So in this case...

     \sec^4 \theta -2\sec^2 \theta tan^2 \theta +tan^4 \theta

    becomes...
    (\sec^2 \theta)^2 - 2\sec^2 \theta tan^2 \theta +(tan^2 \theta)^2

    which is in the form...
    x^2 - 2xy + y^2

    Which is equivalent to the form...
    (x-y)^2

    Therefore the equation can be re-written as...
    (sec^2 \theta -tan^2 \theta)^2

    And then I go on from there?

    How did you recognise that the original equation is the same as the form...(x-y)^2 = x^2 - 2xy + y^2...???

    Are there any other forms that I should be aware of? Or is there a "cheat sheet" of all the kind of forms that I should be familiar with like 'Completing the Square', 'The Difference of Two Squares' etc?

    Thanks again for the help Zacken!!


    (Original post by RDKGames)
    Factorise it. Let x=\sec^2 \theta if you have to and then treat tan as some constant number. Then apply some identities to the bracket to see if it simplifies.

    This is like a simple case of a^2-2ab+b^2=(a-b)^2
    Thank you RDKGames!! I'm slowly getting there with this one
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    (Original post by Philip-flop)
    Therefore the equation can be re-written as...
    (sec^2 \theta -tan^2 \theta)^2

    And then I go on from there?
    Well done.

    Once you see \sec^2 x and \tan^2 in the same place, you should automatically write down 1 + \tan^2 x = \sec^2 x and then stare at it and re-arrange it and plug it in and use it wherever you can. What do you get from there?

    Whenever you see \cot^2 x anywhere, immediately write down 1 + \cot^2 x = \csc^2 x. Especially if \csc^2 x is involved.

    Whenever you see \sin^2 x or \cos^2 x, write down the identity connecting those two. \sin^2 x + \cos^2 x. Even if you think they won't be useful. Write them down next to the question and stare at them for two minutes.

    So, in this case, can you see what to do?
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    (Original post by Philip-flop)
    How did you recognise that the original equation is the same as the form...(x-y)^2 = x^2 - 2xy + y^2...???

    Are there any other forms that I should be aware of? Or is there a "cheat sheet" of all the kind of forms that I should be familiar with like 'Completing the Square', 'The Difference of Two Squares' etc?

    Thanks again for the help Zacken!!
    Hmmm, this is something that you're going to have to get used to and spot with lots of practice. You should definitely know

    (x-y)(x+y) = x^2 - y^2 off the top of your head.

    (x-y)^2 = x^2 -2xy + y^2 should also always be something you should be intimately familiar with.

    (x+y)^2 = x^2 + 2xy + y^2 same as above for this one.

    Other than that, I don't think there's really anything else you should be extremely familiar with. Just keep practicing. Maybe start your own cheat sheet? Add on all the basic identities, factorising forms, tricks that come up extremely often that you keep forgetting and then whenever you're stuck on a problem, stare at the sheet for a bit. You'll get used to it eventually.
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    (Original post by Philip-flop)
    Thank you RDKGames!! I'm slowly getting there with this one
    No problem.

    It is best to remember expansions for (a+b)(a-b), (a+b)^2 and (a-b)^2 as then you can pick up on these patterns fairly quickly by inspection like Zacken and I did.
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    (Original post by Zacken)
    Well done.

    Once you see \sec^2 x and \tan^2 in the same place, you should automatically write down 1 + \tan^2 x = \sec^2 x and then stare at it and re-arrange it and plug it in and use it wherever you can. What do you get from there?

    Whenever you see \cot^2 x anywhere, immediately write down 1 + \cot^2 x = \csc^2 x. Especially if \csc^2 x is involved.

    Whenever you see \sin^2 x or \cos^2 x, write down the identity connecting those two. \sin^2 x + \cos^2 x. Even if you think they won't be useful. Write them down next to the question and stare at them for two minutes.

    So, in this case, can you see what to do?
    Yes!! I think I need to get into the habit of writing these Trig Identities down next to my workings so it's easier to think about the next step I should be taking.

    Ok. so from...
    (\sec^2 \theta - tan^2 \theta)^2

    I can see that the Trig Identity...  1+tan^2 \theta = \sec^2 \theta... can be rearranged to give... \sec^2 \theta - tan^2 \theta = 1

    Therefore...
    (\sec^2 \theta - tan^2 \theta)^2

    Becomes...
    (1)^2

    Which simiplified is just...
    1
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    (Original post by Philip-flop)
    Yes!! I think I need to get into the habit of writing these Trig Identities down next to my workings so it's easier to think about the next step I should be taking.
    I think that would really help.

    Which simiplified is just... 1
    Yep, well done! By the way, don't feel obliged to LaTeX everything, I'd rather a fast reply than a LaTeX one.
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    (Original post by Zacken)
    Hmmm, this is something that you're going to have to get used to and spot with lots of practice. You should definitely know

    (x-y)(x+y) = x^2 - y^2 off the top of your head.

    (x-y)^2 = x^2 -2xy + y^2 should also always be something you should be intimately familiar with.

    (x+y)^2 = x^2 + 2xy + y^2 same as above for this one.

    Other than that, I don't think there's really anything else you should be extremely familiar with. Just keep practicing. Maybe start your own cheat sheet? Add on all the basic identities, factorising forms, tricks that come up extremely often that you keep forgetting and then whenever you're stuck on a problem, stare at the sheet for a bit. You'll get used to it eventually.
    Thank you. I definitely do need to familiarise myself with those equations with their factorising equivalents because when I see one or the other it's like I'm seeing a complete different equation, whereas I should see be able to spot that they are the same thing.

    Thank you so much !!! You always manage to clear these things up!

    (Original post by RDKGames)
    No problem.

    It is best to remember expansions for (a+b)(a-b), (a+b)^2 and (a-b)^2 as then you can pick up on these patterns fairly quickly by inspection like Zacken and I did.
    Yes it will definitely take me a little bit of time to get used to until I'm able to spot those expansions and immediately think of their factorising equivalent.

    Really appreciate your help. I can't thank you enough!
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    Zacken and RDKGames

    Do you two know of any Maths books that you would recommend for someone like me? I need something that is "dumbed down" for beginners like me as I feel like there are lots of gaps in my knowledge. I just want to stop being such a nonsense and posting of TSR all the time
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    (Original post by Philip-flop)
    Zacken and RDKGames

    Do you two know of any Maths books that you would recommend for someone like me? I need something that is "dumbed down" for beginners like me as I feel like there are lots of gaps in my knowledge. I just want to stop being such a nonsense and posting of TSR all the time
    Not really. C3/C4 picks up from from the content of C1/C2 so you might want to go over those 2. As far as mathematical dexterity and quickness goes with tricks, it's just practice really.
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    (Original post by RDKGames)
    Not really. C3/C4 picks up from from the content of C1/C2 so you might want to go over those 2. As far as mathematical dexterity and quickness goes with tricks, it's just practice really.
    Yes, I think once I've finished working through the Edexcel C3 Modular Maths Book I will revisit a lot of the topics from C1 and C2 before I move on to learning the content for C4
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    Seriously why am I always stuck??...

    Click to enlarge question below...
    Name:  C3 - Exe6D Q3.png
Views: 39
Size:  2.2 KB

    For part (a) So far I've got...

    \cot \theta = - \sqrt{3}

     \cot^2 \theta = -3

    \mathrm{cosec} ^2 \theta -1 =-3

    \mathrm{cosec} ^2 \theta =-2

     sin^2 \theta = \frac{1}{-2}

    But then from there on I know my answer is going to be incorrect
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    (Original post by Philip-flop)
    Seriously why am I always stuck??...

    Click to enlarge question below...
    Name:  C3 - Exe6D Q3.png
Views: 39
Size:  2.2 KB

    For part (a) So far I've got...

    \cot \theta = - \sqrt{3}

     \cot^2 \theta = -3
    This bit is incorrect, you've squared both sides, so: (\cot \theta)^2 = (-\sqrt{3})^2 = -\sqrt{3}\times -\sqrt{3} = +3 but the rest of your method is v. good
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    (Original post by Philip-flop)
    Seriously why am I always stuck??...

    Click to enlarge question below...
    Name:  C3 - Exe6D Q3.png
Views: 39
Size:  2.2 KB

    For part (a) So far I've got...

    \cot \theta = - \sqrt{3}

     \cot^2 \theta = -3

    \mathrm{cosec} ^2 \theta -1 =-3

    \mathrm{cosec} ^2 \theta =-2

     sin^2 \theta = \frac{1}{-2}

    But then from there on I know my answer is going to be incorrect
    Second step. Squaring -\sqrt{3} would give you +3.
 
 
 
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