Maths C3 - Trigonometry... Help??

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    (Original post by Philip-flop)
    I had a feeling that would be the case


    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
    You really shouldn't be spending more than 5 minutes on each of these let alone 30. If you cannot do a question within 10 mins then post it here, or move onto a different one, no point wasting too much time on a single question.
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    (Original post by RDKGames)
    You really shouldn't be spending more than 5 minutes on each of these let alone 30. If you cannot do a question within 10 mins then post it here, or move onto a different one, no point wasting too much time on a single question.
    I know I'm probably wasting too much time but I'm sure you guys must get fed up with my constant cries for help!

    It's crazy how I've been stuck on Exercise 6D questions for the last two days and still haven't made much progress (I've still got another 4 questions to do!!). This chapter is so close to defeating me
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    (Original post by Philip-flop)
    I had a feeling that would be the case


    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
    It'd be useful if you let us know which questions exactly you're stuck on so that any recurring mistakes will be obvious.
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    Ok so now my basic Maths knowledge is all over the place again..

    Now I can't even seem to do the inverse of squaring (square rooting) equations

    say for this example...

    \sec^2k - 2 \sec k = 1

    Edit: Sorry that's probably not the best example

    but for something that is like...

     x^2 + 4x = 16 <<<What would be the process?

    would it become...
     x + \sqrt 4x = \sqrt 16

     x+ 2x^\frac{1}{2} = 4
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    (Original post by Philip-flop)
    Ok so now my basic Maths knowledge is all over the place again..

    Now I can't even seem to do the inverse of squaring (square rooting) equations

    say for this example...

    \sec^2k - 2 \sec k = 1

    Edit: Sorry that's probably not the best example

    but for something that is like...

     x^2 + 4x = 16 <<<What would be the process?

    would it become...
     x + \sqrt 4x = \sqrt 16
    Oh dear, that is a completely wrong approach, I'm sad to say.

    When you say "whatever you do to one side, you must do the same to the other" you literally do JUST that. It does NOT mean you do the operation on each individual term on both sides.

    So from x^2+4x=16

    Square root both sides you get \sqrt{x^2+4x}=\sqrt{16}

    I'm not sure why you're applying it to these examples, but that is how you'd square root both sides.
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    Ok so the question I'm stuck on involves finding  \sec k ...

    The original equation is...
     tan^2 k = 2 \sec k

    what I've done is...

     \sec^2 - 1 = 2 \sec k
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    (Original post by Philip-flop)
    Ok so the question I'm stuck on involves finding  \sec k ...

    The original equation is...
     tan^2 k = 2 \sec k

    what I've done is...

     \sec^2 - 1 = 2 \sec k
    subtract 2sec(k) from both sides and solve the quadratic in sec(k).
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    (Original post by RDKGames)
    Oh dear, that is a completely wrong approach, I'm sad to say.

    When you say "whatever you do to one side, you must do the same to the other" you literally do JUST that. It does NOT mean you do the operation on each individual term on both sides.

    So from x^2+4x=16

    Square root both sides you get \sqrt{x^2+4x}=\sqrt{16}

    I'm not sure why you're applying it to these examples, but that is how you'd square root both sides.
    I'm so stupid, I know this!! I think that's enough Maths for one day before I start embarrassing myself even more
    It's these damn Trig Indentity questions that are sucking the knowledge out of me I swear.
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    (Original post by Philip-flop)
    Ok so the question I'm stuck on involves finding  \sec k ...

    The original equation is...
     tan^2 k = 2 \sec k

    what I've done is...

     \sec^2 - 1 = 2 \sec k
    So move 2\sec k to the LHS and complete the square.
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    (Original post by Philip-flop)

    Edit: Sorry that's probably not the best example

    but for something that is like...

     x^2 + 4x = 16 <<<What would be the process?

    would it become...
     x + \sqrt 4x = \sqrt 16

     x+ 2x^\frac{1}{2} = 4
    It seems like you're missing some basic algebra skills, which may be one reason you're finding this topic so hard.

     x^2 + 4x = 16

    This is a quadratic equation. If you move the 16 to the other side, it should be in a form you recognise.

    The way you attempted to square root both sides was a serious algebra mistake that you cannot make at A Level. You really need to make sure you understand your error so you don't make it again. Feel free to ask if you can't see what you did wrong.
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    (Original post by RDKGames)
    So move 2\sec k to the LHS and complete the square.
    Ok so I managed to do...
     \sec^2 k - 2\sec k = 1

     (\sec k - 1)^2 -1 =1

     (\sec k - 1)^2 =2

     \sec k - 1 = \sqrt 2

     \sec k = 1 +\sqrt 2

    Thank you!! How did you know that I needed to complete the square as the next step?
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    (Original post by notnek)
    It seems like you're missing some basic algebra skills, which may be one reason you're finding this topic so hard.

     x^2 + 4x = 16

    This is a quadratic equation. If you move the 16 to the other side, it should be in a form you recognise.

    The way you attempted to square root both sides was a serious algebra mistake that you cannot make at A Level. You really need to make sure you understand your error so you don't make it again. Feel free to ask if you can't see what you did wrong.
    Yes it's because I couldn't find a decent example to show why I was confused. But I will admit that there are a lot of gaps in my basic algebra knowledge!! What do you recon I should do to fill these gaps as I don't know where to start?
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    (Original post by Philip-flop)
    Ok so I managed to do...
     \sec^2 k - 2\sec k = 1

     (\sec k - 1)^2 -1 =1

     (\sec k - 1)^2 =2

     \sec k - 1 = \sqrt 2

     \sec k = 1 +\sqrt 2

    Thank you!! How did you know that I needed to complete the square as the next step?
    Because I mentally moved it to the LHS and recognised that there are no two numbers which multiply to make up -1 and add to make -2. Just a simple deduction by inspection.
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    (Original post by Philip-flop)
    Yes it's because I couldn't find a decent example to show why I was confused. But I will admit that there are a lot of gaps in my basic algebra knowledge!! What do you recon I should do to fill these gaps as I don't know where to start?
    C1
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    (Original post by Naruke)
    C1
    Yeah I think I will revisit C1 again after I've finished the rest of the C3 book tbh. It seems like the break I had from Maths over the summer has been taking its toll on me so is taking me time to get back into the swing of things again (including the very basics).
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    Name:  C3 EXE6D Q10.png
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    So I'm having trouble with expressing b in terms of a.
    I know I have to try and make  4\sec x the same as  cos x

    So what I originally was going to do was start getting rid of the coefficient of sec by dividing by 4 which I soon realised is the wrong method

    So then I looked at it from an equation perspective...  4 \sec x = cos x

    In which I have to make the LHS equal the RHS. Or will it be easier to take the RHS and try to make it equal the LHS??

    But I'm still stuck
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    (Original post by Philip-flop)
    Name:  C3 EXE6D Q10.png
Views: 26
Size:  2.3 KB

    So I'm having trouble with expressing b in terms of a.
    I know I have to try and make  4\sec x the same as  cos x

    So what I originally was going to do was start getting rid of the coefficient of sec by dividing by 4 which I soon realised is the wrong method

    So then I looked at it from an equation perspective...  4 \sec x = cos x

    In which I have to make the LHS equal the RHS. Or will it be easier to take the RHS and try to make it equal the LHS??

    But I'm still stuck
    a/4=secx, now 1/b=secx, can you now find a in terms of b from this?
    Or you can simply multiply a and b together.
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    (Original post by Philip-flop)
    Name:  C3 EXE6D Q10.png
Views: 26
Size:  2.3 KB

    So I'm having trouble with expressing b in terms of a.
    I know I have to try and make  4\sec x the same as  cos x

    So what I originally was going to do was start getting rid of the coefficient of sec by dividing by 4 which I soon realised is the wrong method

    So then I looked at it from an equation perspective...  4 \sec x = cos x

    In which I have to make the LHS equal the RHS. Or will it be easier to take the RHS and try to make it equal the LHS??

    But I'm still stuck
    \displaystyle a=4\sec(x)=4(\frac{1}{\cos(x)})=  4(\frac{1}{b}) and proceed from there
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    (Original post by B_9710)
    a/4=secx, now 1/b=secx, can you now find a in terms of b from this?
    Ok so what I've done is...

     a = 4 \sec x

     \frac{a}{4} = \sec x

     \frac{a}{4} = \frac{1}{cos x}

    Flip the fractions on both sides to give...

     \frac{4}{a} = cos x

     \frac{4}{a} = b
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    (Original post by Philip-flop)
    Ok so what I've done is...

     a = 4 \sec x

     \frac{a}{4} = \sec x

     \frac{a}{4} = \frac{1}{cos x}

    Flip the fractions on both sides to give...

     \frac{4}{a} = cos x

     \frac{4}{a} = b
    That's right.
 
 
 
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