Maths C3 - Trigonometry... Help??

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    In my above post I'm assuming that it's an equation you're supposed to solve for \theta. If by making the LHS equal the RHS you mean proving it as an identity then that's not possible because it's not an identity.
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    (Original post by Philip-flop)
    An identity with a '2' in it? I've never come across that yet
    Well that was a really mess up with the latex lol. I meant an identity with a \cot^2 in it.
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    (Original post by RDKGames)
    Are you sure that is the question? The two are not identical.
    OMG, my brain really is fried today!! I must have been looking at the answers to a complete different question

    It's definitely one of those days where I should have just crawled back into bed as soon as I got home from work Seriously FML!!

    Sorry for wasting everyone's time. I will refrain from posting for the rest of the night.
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    (Original post by Philip-flop)
    OMG, my brain really is fried today!! I must have been looking at the answers to a complete different question

    It's definitely one of those days where I should have just crawled back into bed as soon as I got home from work Seriously FML!!

    Sorry for wasting everyone's time. I will refrain from posting for the rest of the night.
    Relax
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    These Trig Identities are seriously testing my patience!

    I've tried absolutely everything with this question to the point where I've started making up my own rules

    Name:  C3 - EXE6D Q(7) Trig Identities.png
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    So I'm pretty certain I can start off by using the Trig Identity 1+ tan^2 \theta = \sec ^2 \theta

    to give me...
    3tan^2 \theta +4(1+tan^2 \theta) = 5

    Then I expand the brackets and re-arrange the equation to give me...
     7tan^2 \theta =1 <<<have I done something wrong already?
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    (Original post by Philip-flop)
    These Trig Identities are seriously testing my patience!

    I've tried absolutely everything with this question to the point where I've started making up my own rules

    Name:  C3 - EXE6D Q(7) Trig Identities.png
Views: 28
Size:  2.0 KB

    So I'm pretty certain I can start off by using the Trig Identity 1+ tan^2 \theta = \sec ^2 \theta

    to give me...
    3tan^2 \theta +4(1+tan^2 \theta) = 5

    Then I expand the brackets and re-arrange the equation to give me...
     7tan^2 \theta =1 <<<have I done something wrong already?
    You're correct. Why is it wrong??
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    (Original post by RDKGames)
    You're correct. Why is it wrong??
    Because I feel like I get stuck from there on

    I know I can use the Identity  tan \theta = \frac{sin \theta}{cos \theta} but then I have no idea how to solve just for  sin \theta Things just get too messy
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    (Original post by Philip-flop)
    Because I feel like I get stuck from there on

    I know I can use the Identity  tan \theta = \frac{sin \theta}{cos \theta} but then I have no idea how to solve just for  sin \theta
    So you got the value of 7\tan^2(\theta)=1 which means that 7\sin^2(\theta)=\cos^2(\theta) (mult. both sides by cosine squared after expressing tan in terms of sine and cosine) then express cosine squared in terms of sine squared then square root sine term and get the correct value for an obtuse angle.
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    (Original post by RDKGames)
    So you got the value of 7\tan^2(\theta)=1 which means that 7\sin^2(\theta)=\cos^2(\theta) (mult. both sides by cosine squared after expressing tan in terms of sine and cosine) then express cosine squared in terms of sine squared then square root sine term and get the correct value for an obtuse angle.
    Ok so from...
     7sin^2 \theta = cos^2 \theta

    This gives me...
     7sin^2 \theta = 1- sin^2 \theta

    Then add sin^2(theta) to both sides? to give...
     8sin^2 \theta = 1

     sin^2 \theta = \frac{1}{8}

     sin \theta = \sqrt \frac{1}{8}

    To which I end up having to rationalise the denominator?
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    (Original post by Philip-flop)
    Ok so from...
     7sin^2 \theta = cos^2 \theta

    This gives me...
     7sin^2 \theta = 1- sin^2 \theta

    Then add sin^2(theta) to both sides? to give...
     8sin^2 \theta = 1

     sin^2 \theta = \frac{1}{8}

     sin \theta = \sqrt \frac{1}{8}

    To which I end up having to rationalise the denominator?
    Correct working, and you don't have to rationalise it, but remember that square rooting gives \pm the answer and you need to state the reason for your choice.
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    (Original post by RDKGames)
    Correct working, and you don't have to rationalise it, but remember that square rooting gives \pm the answer and you need to state the reason for your choice.
    Oh yeah of course!! \theta only appears between 90 and 180 (in the 2nd quadrant) when sin is +ve

    Therefore...
     sin \theta \not = - \frac {\sqrt 2}{4}

    so...
     sin \theta = + \frac {\sqrt 2}{4}

    Thanks again for your help!! I can't even begin to explain how much I appreciate it! Maybe I should donate money to you haha. Or pay you to tutor me :P
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    (Original post by Philip-flop)
    Oh yeah of course!! \theta only appears between 90 and 180 (in the 2nd quadrant) when sin is +ve

    Therefore...
     sin \theta \not = - \frac {\sqrt 2}{4}

    so...
     sin \theta = + \frac {\sqrt 2}{4}

    Thanks again for your help!! I can't even begin to explain how much I appreciate it! Maybe I should donate money to you haha. Or pay you to tutor me :P
    Haha no problem, well done
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    Ok so I realise I absolutely suck at trigonometry as I'm stuck on yet another question ...

    Q) Solve the equation for the following interval...
    Name:  C3 Trignometry EXE2D Q8d.png
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    Am I right in thinking that the trig identity... 1 + \cot^2 \theta = \mathrm{cosec}^2 \theta... can be rearranged to give...
    1- \mathrm{cosec}^2 \theta = -\cot ^2 \theta

    So for this question I would do...
     \cot \theta = 1-\mathrm{cosec}^2 \theta

     \cot \theta = - \cot ^2 \theta
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    (Original post by Philip-flop)
    Ok so I realise I absolutely suck at trigonometry as I'm stuck on yet another question ...

    Q) Solve the equation for the following interval...
    Name:  C3 Trignometry EXE2D Q8d.png
Views: 22
Size:  822 Bytes

    Am I right in thinking that the trig identity... 1 + \cot^2 \theta = \mathrm{cosec}^2 \theta... can be rearranged to give...
    1- \mathrm{cosec}^2 \theta = -\cot ^2 \theta

    So for this question I would do...
     \cot \theta = 1-\mathrm{cosec}^2 \theta

     \cot \theta = - \cot ^2 \theta
    Okay so you cannot divide otherwise you'd be losing solutions.

    From moving RHS onto LHS, you get \cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for \theta in the given range.
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    (Original post by RDKGames)
    Okay so you cannot divide otherwise you'd be losing solutions.

    From moving RHS onto LHS, you get \cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for \theta in the given range.
    You'll also notice why \cot\theta \not \equiv \dfrac{1}{\tan\theta} from solving this equation.
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    (Original post by RDKGames)
    Okay so you cannot divide otherwise you'd be losing solutions.

    From moving RHS onto LHS, you get \cot^2 \theta + \cot \theta = 0 at which point you can factorise LHS and solve two equations for \theta in the given range.
    Ok I'm not sure how I still ended up losing 2 of the solutions

    This is what I got...
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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.
    Yeah because it'll give you no solution right? How do i work around that?
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    (Original post by Philip-flop)
    Ok I'm not sure how I still ended up losing 2 of the solutions

    This is what I got...
    Correct solutions.

    That is the pitfall you made as the question aims to throw people off, and IrrationalRoot pointed out

    \cot \theta \not\equiv \frac{1}{\tan \theta}

    If you turn \cot \theta \equiv \frac{\cos \theta}{\sin \theta} = 0 then you can see if \cos \theta = 0 then your equation still holds and you get more solutions.
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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.
    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.
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    (Original post by RDKGames)
    Correct solutions.

    That is the pitfall you made as the question aims to throw people off, and IrrationalRoot pointed out

    \cot \theta \not\equiv \frac{1}{\tan \theta}

    If you turn \cot \theta \equiv \frac{\cos \theta}{\sin \theta} = 0 then you can see if \cos \theta = 0 then your equation still holds and you get more solutions.
    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?
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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?
    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.
 
 
 
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