Maths C3 - Trigonometry... Help??

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    (Original post by Philip-flop)
    Seriously why am I always stuck??...

    Click to enlarge question below...
    Attachment 580786

    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
    Not saying you haven't but you should at least check your working before asking for help. Putting as much effort in as you can before asking for help is much better, since you'll actually learn that way. If you're doing this already then that's good and you can ignore this, but I would think that having checked your working you'd realise that (-\sqrt{3})^2 \not= -3.

    Anyway, your method is fine but alternatively you can just find \theta by rewriting the original equation as \tan\theta=-\dfrac{1}{\sqrt{3}}.
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    (Original post by Zacken)
    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
    Oh dear. Such a rookie mistake as well!!
    Not sure how I slipped up on that!! I must have just cancelled the square root on the RHS and then squared the LHS.
    I think that's enough Maths for one day :/

    Thanks for pointing that one out though!

    (Original post by RDKGames)
    Second step. Squaring -\sqrt{3} would give you +3.
    Doh, I'm so stupid. Thanks for not laughing at me
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    (Original post by IrrationalRoot)
    Not saying you haven't but you should at least check your working before asking for help. Putting as much effort in as you can before asking for help is much better, since you'll actually learn that way. If you're doing this already then that's good and you can ignore this, but I would think that having checked your working you'd realise that (-\sqrt{3})^2 \not= -3.

    Anyway, your method is fine but alternatively you can just find \theta by rewriting the original equation as \tan\theta=-\dfrac{1}{\sqrt{3}}.
    Yeah I feel really embarrassed by that silly mistake now
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    (Original post by Philip-flop)
    Yeah I feel really embarrassed by that silly mistake now
    Hi ,
    If u wanna become a master at trigonometry (im just thinking...)
    Learn all about visual perspective of trigonometric functions,
    Things like trig graphs and unit circles help a lot in the long run compared to cast diagrams (but then thats only me...)
    I suggest khanacademy.org and patrickjmt
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    (Original post by Philip-flop)
    Seriously why am I always stuck??...

    Click to enlarge question below...
    Attachment 580786

    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
    An alternate method:

    Ignore the negative and say \displaystyle \cot \theta = \frac{\sqrt{3}}{1} = \frac{adj}{opp}.

    Then draw a right-angled triangle and use this to find \sin \theta and use the cast diagram/graph to get the sign of the answer.
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    (Original post by me2*)
    Hi ,
    If u wanna become a master at trigonometry (im just thinking...)
    Learn all about visual perspective of trigonometric functions,
    Things like trig graphs and unit circles help a lot in the long run compared to cast diagrams (but then thats only me...)
    I suggest khanacademy.org and patrickjmt
    This is true. If you really want to understand trigonometry the best way is learning the unit circle definition (which is really the only definition; the right triangle definition has its limitations) and using the unit circle to solve problems. If you visualise the trig functions via the unit circle you really can't get caught out.
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    (Original post by notnek)
    An alternate method:

    Ignore the negative and say \displaystyle \cot \theta = \frac{\sqrt{3}}{1} = \frac{adj}{opp}.

    Then draw a right-angled triangle and use this to find \sin \theta and use the cast diagram/graph to get the sign of the answer.
    I will definitely need to look into more of using the right angled triangle method!
    (Original post by IrrationalRoot)
    This is true. If you really want to understand trigonometry the best way is learning the unit circle definition (which is really the only definition; the right triangle definition has its limitations) and using the unit circle to solve problems. If you visualise the trig functions via the unit circle you really can't get caught out.
    Yes this way is definitely more visual so might help me out quite a bit.
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    Ok so for part (b) where have I gone wrong?...

    Name:  C3 - Exe6D Q3.png
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    \cot \theta =-\sqrt {3}

     \cot^2 \theta = 3

     tan^2 \theta =\frac{1}{3}

    \sec^2 \theta -1=\frac{1}{3}

    \sec^2 \theta = \frac{4}{3}

    cos^2 \theta = \frac{1}{\frac{4}{3}}

     cos^2 \theta = \frac{3}{4}

     cos \theta = \sqrt {\frac{3}{4}}

     cos \theta = \frac{\sqrt {3} }{2}

    I know the answer is meant to be..  cos \theta = -\frac{\sqrt {3} }{2}
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    (Original post by Philip-flop)
    Ok so for part (b) where have I gone wrong?...
    When you took:

    \cos ^{ 2 }{ \theta  } =\frac { 3 }{ 4 }

    You forgot the plus or minus sign when using the square root.

    \cos { \theta  } =\pm \frac { \sqrt { 3 }  }{ 2 }

    Given that 90^{\circ}<\theta<180^{\circ}, your answer has to be:

    \cos { \theta } =- \frac { \sqrt { 3 } }{ 2 }
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    (Original post by Philip-flop)
    Ok so for part (b) where have I gone wrong?...


     cos^2 \theta = \frac{3}{4}

     cos \theta = \sqrt {\frac{3}{4}}

    I know the answer is meant to be..  cos \theta = -\frac{\sqrt {3} }{2}
    When it comes to square rooting, the solution is \cos \theta = \pm \frac{\sqrt{3}}{2}

    It cannot be the positive because then the angle is out of the given domain 90 < \theta < 180


    You might find it useful to get yourself a sketch like this. Draw the function from 0 to 360. Then highlight yourself the restricted region. When you get something like \cos \theta = \pm \frac{\sqrt{3}}{2}, draw the lines y=\frac{\sqrt{3}}{2} (green line) and y=-\frac{\sqrt{3}}{2} (black line). Whichever line hits the function WITHIN the highlighted region, that's the correct solution.

    Of course, you may also find that you don't even need to sketch the lines because the range (y-values) you're given for cosine is strictly between 0 and -1, so it's going to be anything with a minus in front.

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    Since you were working with CAST diagrams earlier, remember that the quadrant where 90 < theta < 180 is precisely the second quadrant, the "S" quadrant where sine is positive and cos and tan are negative. So you should be picking the negative root.
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    (Original post by Zacken)
    Since you were working with CAST diagrams earlier, remember that the quadrant where 90 < theta < 180 is precisely the second quadrant, the "S" quadrant where sine is positive and cos and tan are negative. So you should be picking the negative root.
    Thanks Zacken!! It makes sense to me now. Drawing the cast diiagram out and writing down  90&lt; \theta &lt;180 made me realise that cos only appears in the second quadrant when cos is -ve
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    (Original post by Philip-flop)
    Thanks Zacken!! It makes sense to me now. Drawing the cast diiagram out and writing down  90&lt; \theta &lt;180 made me realise that cos only appears in the second quadrant when cos is -ve
    Yep!
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    (Original post by Philip-flop)
    Thanks Zacken!! It makes sense to me now. Drawing the cast diiagram out and writing down  90&lt; \theta &lt;180 made me realise that cos only appears in the second quadrant when cos is -ve
    https://www.youtube.com/watch?v=eucID63O8D4
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    (Original post by Philip-flop)
    Ok so for part (b) where have I gone wrong?...

    Name:  C3 - Exe6D Q3.png
Views: 27
Size:  2.2 KB

    \cot \theta =-\sqrt {3}

     \cot^2 \theta = 3

     tan^2 \theta =\frac{1}{3}

    \sec^2 \theta -1=\frac{1}{3}

    \sec^2 \theta = \frac{4}{3}

    cos^2 \theta = \frac{1}{\frac{4}{3}}

     cos^2 \theta = \frac{3}{4}

     cos \theta = \sqrt {\frac{3}{4}}

     cos \theta = \frac{\sqrt {3} }{2}

    I know the answer is meant to be..  cos \theta = -\frac{\sqrt {3} }{2}
    It's worth pointing out that you're making an awful lot of extra work for yourself here, as well as inviting the possibility of making more algebraic slip-ups!

    You know (or should do!) that cot X = cos X / sin X for any X; the question tells you what cot X is; and in part (a) you worked out what sin X was. So now you can rearrange this equation to get cos X in terms of known quantities
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    Here I am again... stuck.

    How do I make the LHS equal the RHS...
    \mathrm{cosec}^2 \theta +1 = 3\cot \theta

    The book suggests that the next step should be...
     \frac{1}{cos^2\theta} (\frac{cos^2 \theta}{sin^2 \theta} - cos^2 \theta) ... Which means the step after that is...  \frac{1}{sin^2 \theta} - 1

    Surely the -1 doesn't make it the same as the original equation??
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    Anyone? Please...
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    (Original post by Philip-flop)
    Here I am again... stuck.

    How do I make the LHS equal the RHS...
    \mathrm{cosec}^2 \theta +1 = 3\cot \theta

    The book suggests that the next step should be...
     \frac{1}{cos^2\theta} (\frac{cos^2 \theta}{sin^2 \theta} - cos^2 \theta) ... Which means the step after that is...  \frac{1}{sin^2 \theta} - 1

    Surely the -1 doesn't make it the same as the original equation??
    Well the goal with any trig equation is to get the equation in terms of just one trig function. So keep that in mind when I give you the hint that you (should) know an identity with \cosec^2 in it.
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    (Original post by Philip-flop)
    Here I am again... stuck.

    How do I make the LHS equal the RHS...
    \mathrm{cosec}^2 \theta +1 = 3\cot \theta

    The book suggests that the next step should be...
     \frac{1}{cos^2\theta} (\frac{cos^2 \theta}{sin^2 \theta} - cos^2 \theta) ... Which means the step after that is...  \frac{1}{sin^2 \theta} - 1

    Surely the -1 doesn't make it the same as the original equation??
    Are you sure that is the question? The two are not identical.
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    (Original post by IrrationalRoot)
    Well the goal with any trig equation is to get the equation in terms of just one trig function. So keep that in mind when I give you the hint that you (should) know an identity with \cosec^2 in it.
    An identity with a '2' in it? I've never come across that yet
 
 
 
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