Maths C3 - Trigonometry... Help??

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    (Original post by Philip-flop)
    Oh right so I should have known that... \cot \theta \equiv \frac{1}{\tan \theta} ... isn't actually an identity. I seriously thought it was! Why are the books so misleading at times?

    Ok so now that I know. I should have done this...

     \cot \theta = 0

     \frac{1}{tan \theta} = 0

     \frac{cos \theta}{sin \theta} = 0

    Then I times both sides by sin \theta to give...

     cos \theta = 0

    Then just work out the other solutions from there?
    Question for my own interest (not meant to be patronising and I don't know the answer to it) but how easily do you feel that you give up trying to solve a question, in particular for when you post here?

    This is to kind of understand A. How helpful this is and B. Whether you can change your approach to questions slightly. I
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    (Original post by IrrationalRoot)
    You'll also notice why \cot\theta \not \equiv \dfrac{1}{\tan\theta} from solving this equation.
    (Original post by RDKGames)
    I find it absurd how so many place say that it is an IDENTITY, when this question is a perfect example of why it is not.
    It is an identity. Do you claim that 1 + \tan^2 \theta = \sec^2 \theta isn't an identity because it fails for \theta = \frac{\pi}{2}?
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    (Original post by RDKGames)
    Essentially yeah, but your second line wouldn't be correct because if I were to multiply both sides by \tan \theta then I would be left with 1=0\cdot \tan \theta = 0 \Rightarrow 1=0 which doesn't make sense.
    Again, you made the same mistake by assuming \cot \theta \equiv \frac{1}{\tan \theta}. If you ignore line 2, then that working would be fine.

    Also it would see more straight forward if from your fractorised form you went:

    \frac{\cos \theta}{\sin \theta}(1+\cot \theta)=0 \Rightarrow \cos \theta (1+\cot \theta)=0

    From multiplying both sides by sine.
    Thank you so much! I never would have understood what happened to those missing solutions if you didn't help me.

    Also, thanks IrrationalRoot for pointing it out too

    (Original post by SeanFM)
    Question for my own interest (not meant to be patronising and I don't know the answer to it) but how easily do you feel that you give up trying to solve a question, in particular for when you post here?

    This is to kind of understand A. How helpful this is and B. Whether you can change your approach to questions slightly. I
    In all honesty I must have spent a good hour trying to solve that one question I'm self-teaching Maths so when I'm stuck, I really get stuck! Then I start to lose faith in myself. Doesn't help that sometimes I feel like there are huge gaps in my knowledge. Also sometimes the Edexcel Endorsed textbook doesn't explain a particular topic too well, so for an amateur like me it's really difficult to get my head around some things

    Last year I wasted so much time trying to understand some things which could have easily been explained to me from someone else's perspective. Also, if I really struggled to get my head around a topic I would move on without fully understanding, and hoping that I wouldn't get tested on it. :/
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    (Original post by Philip-flop)
    In all honesty I must have spent a good hour trying to solve that one question I'm self-teaching Maths so when I'm stuck, I really get stuck! Then I start to lose faith in myself. Doesn't help that sometimes I feel like there are huge gaps in my knowledge. Also sometimes the Edexcel Endorsed textbook doesn't explain a particular topic too well, so for an amateur like me it's really difficult to get my head around some things
    Do you ever use ExamSolutions? His explanations are excellent and go beyond what a textbook can tell you. There's a tutorial for every A Level topic.

    I wish there was something like ExamSolutions back when I self-taught Further Maths. I used a textbook like you and spent a lot of time asking questions on this forum. Self-teaching can be really tough sometimes so I know how you feel.

    Keep asking questions on TSR and that will definitely help. As you've probably noticed there are many excellent members in this forum who can explain things better than a textbook.
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    (Original post by notnek)
    Do you ever use ExamSolutions? His explanations are excellent and go beyond what a textbook can tell you. There's a tutorial for every A Level topic.

    I wish there was something like ExamSolutions back when I self-taught Further Maths. I used a textbook like you and spent a lot of time asking questions on this forum. Self-teaching can be really tough sometimes so I know how you feel.

    Keep asking questions on TSR and that will definitely help. As you've probably noticed there are many excellent members in this forum who can explain things better than a textbook.
    Yeah I use ExamSolutions from time to time. I like to work through the examples in the Edexcel Modular Maths Textbook then watch some videos if I still don't fully understand something
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    (Original post by Zacken)
    It is an identity. Do you claim that 1 + \tan^2 \theta = \sec^2 \theta isn't an identity because it fails for \theta = \frac{\pi}{2}?
    \cot\theta and \dfrac{1}{\tan\theta} are not the same function, that's for sure. So whether or not there should be an identity sign between them seems unclear to me. Maybe the definition of an identity is that both sides are equal for all values such that both sides are defined in which case you're correct.
    Notation aside, what I meant to get across is that the two are not the same function.
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    (Original post by Zacken)
    It is an identity. Do you claim that 1 + \tan^2 \theta = \sec^2 \theta isn't an identity because it fails for \theta = \frac{\pi}{2}?
    Also I realised that if the definition of 'f and g are identical' is that 'f(x)=g(x) for all values such that both f and g are defined' then we get something weird: two functions can be identical but not the same. Lol. The most basic example is probably \dfrac{x}{x} and 1.
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    When solving difficult equations like...

     (sec \theta - cos \theta)^2 = tan \theta - sin^2 \theta .... for the interval...  0 \leq \theta \leq \pi

    Out of sin, cos, and tan how do you know which one it'll likely simplify down to?
    I feel like I'm just doing this out of trial and error without an aim
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    (Original post by Philip-flop)
    When solving difficult equations like...

     (sec \theta - cos \theta)^2 = tan \theta - sin^2 \theta .... for the interval...  0 \leq \theta \leq \pi

    Out of sin, cos, and tan how do you know which one it'll likely simplify down to?
    I feel like I'm just doing this out of trial and error without an aim
    Most likely cosine as it dominates most of the equation, but you'd usually have to play around with it and see first.
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    (Original post by Philip-flop)
    When solving difficult equations like...

     (sec \theta - cos \theta)^2 = tan \theta - sin^2 \theta .... for the interval...  0 \leq \theta \leq \pi

    Out of sin, cos, and tan how do you know which one it'll likely simplify down to?
    I feel like I'm just doing this out of trial and error without an aim
    It's not always possible (or realistic) to see a solution right to its end. Here a sensible thing to do would be to expand the brackets. One can see in advance that expanding the brackets simplifies things a lot.
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    (Original post by RDKGames)
    Most likely cosine as it dominates most of the equation, but you'd usually have to play around with it and see first.
    Yes, I see there are actually quite a few cosines in the equation. Thank you

    (Original post by IrrationalRoot)
    It's not always possible (or realistic) to see a solution right to its end. Here a sensible thing to do would be to expand the brackets. One can see in advance that expanding the brackets simplifies things a lot.
    Yeah straight away I noticed that the squared bracket could be expanded because of the form  (a-b)^2 = a^2 -2ab=b^2

    But 10 lines worth of workings I still find myself stuck on this question
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    (Original post by Philip-flop)
    Yes, I see there are actually quite a few cosines in the equation. Thank you


    Yeah straight away I noticed that the squared bracket could be expanded because of the form  (a-b)^2 = a^2 -2ab=b^2

    But 10 lines worth of workings I still find myself stuck on this question
    Squaring out the bracket should make it very easy...
    First term is \sec^2, second term is constant, third term if moved to the right obviously gives a constant and then you're left with a very standard trig equation that you've probably solved plenty of.
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    (Original post by Philip-flop)
    Yes, I see there are actually quite a few cosines in the equation. Thank you


    Yeah straight away I noticed that the squared bracket could be expanded because of the form  (a-b)^2 = a^2 -2ab=b^2

    But 10 lines worth of workings I still find myself stuck on this question
    (\sec \theta - \cos \theta)^2 = \tan \theta - \sin^2 \theta

    \Rightarrow \sec^2 \theta - 2 + \cos^2 \theta = \tan \theta - \sin^2 \theta

    \Rightarrow \sec^2 \theta -2 + (\cos^2 \theta + \sin^2 \theta) = \tan \theta

    Can you see where to go from there? The \sec^2 \theta can be expressed in terms of \tan \theta and the bracket is your base identity. Working can be done in 5 lines max.
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    (Original post by RDKGames)
    (\sec \theta - \cos \theta)^2 = \tan \theta - \sin^2 \theta

    \Rightarrow \sec^2 \theta - 2 + \cos^2 \theta = \tan \theta - \sin^2 \theta

    \Rightarrow \sec^2 \theta -2 + (\cos^2 \theta + \sin^2 \theta) = \tan \theta

    Can you see where to go from there? The \sec^2 \theta can be expressed in terms of \tan \theta and the bracket is your base identity. Working can be done in 5 lines max.
    OMG!! How did I not see the...  cos^2 \theta + sin^2 \theta??? I even had it written in that order!!

    Why are these trig equations so difficult for me to get used to?
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    (Original post by Philip-flop)
    OMG!! How did I not see the...  cos^2 \theta + sin^2 \theta??? I even had it written in that order!!

    Why are these trig equations so difficult for me to get used to?
    Because you don't do them as much before C3, you'll be fine
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    (Original post by RDKGames)
    Because you don't do them as much before C3, you'll be fine
    Thank you. I sure hope so!! I'm beginning to question whether I'm actually capable
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    Seriously on the verge of giving up

    Does anyone have any advice for these type of Trig Identity questions that I keep getting stuck on? There's got to be some systematic way of working through questions because I am just baffled I feel so useless!!
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    (Original post by Philip-flop)
    Seriously on the verge of giving up

    Does anyone have any advice for these type of Trig Identity questions that I keep getting stuck on? There's got to be some systematic way of working through questions because I am just baffled I feel so useless!!
    There is no systematic way, this isn't GCSE maths anymore. You just have to be aware of all your identities and ultimately aim to get your equations in terms on a single trigonometric function, from which point they are easy to solve.
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    (Original post by Philip-flop)
    Seriously on the verge of giving up

    Does anyone have any advice for these type of Trig Identity questions that I keep getting stuck on? There's got to be some systematic way of working through questions because I am just baffled I feel so useless!!
    Just as RDK said, there is no systematic way. If there was, everyone would be coasting through every question.

    I mean, it doesn't get anymore systematic than

    1. Be as familiar as possible (and on the look out for) identities that you know. The main ones being \sin^2+\cos^2=1 and the two corresponding identities obtained by dividing this equation.

    2. The whole goal with any problem with any new notation/function is getting rid of the new notation/function, which is usually achieved by isolating it. In the case of trig equations it's converting the equation into one in terms of just trig function.
    Then it's easy to solve as normal. Keep that it mind and you should be fine.

    And no it's not as black and white as you were thinking where you're supposed to decide what to convert to at the start - this isn't always possible. Just use the identities you know to simplify things and ultimately get the equation in terms of one trig function. I can guarantee the next time you get stuck, it will be because you either didn't apply one of the standard identities you know (purposefully of course) or you didn't make some sort of obvious simplification to the equation such as clearing fractions, expanding brackets etc. Or you just made a silly mistake lol.
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    (Original post by RDKGames)
    There is no systematic way, this isn't GCSE maths anymore. You just have to be aware of all your identities and ultimately aim to get your equations in terms on a single trigonometric function, from which point they are easy to solve.
    I had a feeling that would be the case

    (Original post by IrrationalRoot)
    Just as RDK said, there is no systematic way. If there was, everyone would be coasting through every question.

    I mean, it doesn't get anymore systematic than

    1. Be as familiar as possible (and on the look out for) identities that you know. The main ones being \sin^2+\cos^2=1 and the two corresponding identities obtained by dividing this equation.

    2. The whole goal with any problem with any new notation/function is getting rid of the new notation/function, which is usually achieved by isolating it. In the case of trig equations it's converting the equation into one in terms of just trig function.
    Then it's easy to solve as normal. Keep that it mind and you should be fine.

    And no it's not as black and white as you were thinking where you're supposed to decide what to convert to at the start - this isn't always possible. Just use the identities you know to simplify things and ultimately get the equation in terms of one trig function. I can guarantee the next time you get stuck, it will be because you either didn't apply one of the standard identities you know (purposefully of course) or you didn't make some sort of obvious simplification to the equation such as clearing fractions, expanding brackets etc. Or you just made a silly mistake lol.
    Thank you. I've started getting into the habit of writing trig identities down as I go along so I can spot any simplifications whilst I'm staring at the equation. But why does it seem that these types of questions take me like 30 minutes each? I'm stuck on another question as we speak that's how much I'm struggling
 
 
 
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