Maths C3 - Trigonometry... Help??

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    (Original post by ValerieKR)
    simplify cos(x)sec(x) and put the remaining terms over a cos(x) denominator
    Thank you soooooo much!! I managed to work it out now. Yayyyyy.
    Seriously appreciate your help
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    Another few of my silly questions...

    Q1) With quadrant diagrams is...
    sin Positive in the 1st and 2nd quadrant --- Negative in the 3rd and 4th quadrant
    cos Positive in the 1st and 4th --- Negative in the 2nd and 3rd
    tan Postive in the 1st and 3rd --- Negative in the 2nd and 4th
    ???

    Q2) How come...

     \theta = cos^-^1(\frac{1}{\sqrt{\frac{4}{3}}})

    ...the square root makes cos positive and negative but...

     \theta = tan^-^1(\sqrt[3]{\frac{1}{8}})

    ...with the cube root in this equation tan just remains positive??

    I know that square roots can are +/- but does the same not apply to cube roots?


    Sorry for the ridiculously silly questions
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    (Original post by Philip-flop)
    ...
    If the original question was \cos^2 \theta = \frac{3}{4} then you get \cos \theta = \pm \sqrt{\frac{3}{4}} so you get two different trig equations to solve.

    But with ]tan^3 \theta = \frac{1}{8}, cube rooting gives only \tan \theta = + \sqrt[3]{\frac{1}{8}} which is only one trig equation to solve. I'm not really sure what you're asking tbh, could you post the original question and your working out?
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    (Original post by Philip-flop)

    I know that square roots can are +/- but does the same not apply to cube roots?
    Sorry for the ridiculously silly questions

    Nope.

    Take -1 for instance. Squaring it, or raising it to any even power, will give the result of +1. This is also the case for the number +1 itself, if you raise it to any even power.

    If you were to cube -1, then you would be multiplying it by itself 3 times and you come back to -1 again. Same with any odd power. Now if you apply odd powers to the number +1, then you would just come back to +1 again.

    So; for even powers, all numbers go positive; hence +/- solutions for any equations with square roots. For odd powers, negatives stay negative, and positives stay positive. Therefore odd roots of a negative will be negative, and odd roots of a positive are positive. Hence no +/- in the answer.
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    (Original post by Zacken)
    could you post the original question and your working out?
    Taken from the Edexcel C3 Modular Maths Textbook - Exercise 6C Questions: 5(e) and 5(g)

    Solve, for the values of \theta in the interval 0 \leq \theta \leq 360
    ...

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    (Original post by RDKGames)
    Nope.

    Take -1 for instance. Squaring it, or raising it to any even power, will give the result of +1. This is also the case for the number +1 itself, if you raise it to any even power.

    If you were to cube -1, then you would be multiplying it by itself 3 times and you come back to -1 again. Same with any odd power. Now if you apply odd powers to the number +1, then you would just come back to +1 again.

    So; for even powers, all numbers go positive; hence +/- solutions for any equations with square roots. For odd powers, negatives stay negative, and positives stay positive. Therefore odd roots of a negative will be negative, and odd roots of a positive are positive. Hence no +/- in the answer.
    Oh yeah!! That's it! Thank you so much. I don't know why but I seem to know these things already but when it comes to applying it to questions I panic and forget simple things like that.

    Great explanation. It makes perfect sense now.
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    (Original post by Philip-flop)
    Taken from the Edexcel C3 Modular Maths Textbook - Exercise 6C Questions: 5(e) and 5(g)

    Solve, for the values of \theta in the interval 0 \leq \theta \leq 360
    A bit messy.

    So you have \sec^2\theta = \frac{4}{3} so \cos^2\theta = \frac{3}{4} since you take reciprocals of both sides.

    At which point \cos \theta = \pm \frac{\sqrt3}{2}

    Your tan looks good but it would just be easier to rewrite cube root of 1/8 as 1/2.

    Also I think learning and remembering the general solutions of trigonometric functions in FP1 can make it so much neater for you to solve these types in C3. It's a really short topic, look at it if you can. I feel like the quadrant thing just leaves much more room for error.
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    (Original post by RDKGames)
    Your tan looks good
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    (Original post by RDKGames)
    A bit messy.

    So you have \sec^2\theta = \frac{4}{3} so \cos^2\theta = \frac{3}{4} since you take reciprocals of both sides.

    At which point \cos \theta = \pm \frac{\sqrt3}{2}

    Your tan looks good but it would just be easier to rewrite cube root of 1/8 as 1/2.

    Also I think learning and remembering the general solutions of trigonometric functions in FP1 can make it so much neater for you to solve these types in C3. It's a really short topic, look at it if you can. I feel like the quadrant thing just leaves much more room for error.
    Oh yeah. I didn't think of just taking the reciprocal of both sides since.. sec is the reciprocal of cos

    This is why I will never be goood at Maths
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    Ok, here I go again...

    Solve, for the values of \theta in the interval -180 \leq \theta \leq 180

    How do I even solve this equation?...

     3 \cot \theta =2sin\theta
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    (Original post by Philip-flop)
    Ok, here I go again...

    Solve, for the values of \theta in the interval -180 \leq \theta \leq 180

    How do I even solve this equation?...

     3 \cot \theta =2sin\theta
    Rewrite cot in terms of sin and cos. Get sine squared, get cos squared from that. Boom you got a quadratic in terms of cos.
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    (Original post by RDKGames)
    Rewrite cot in terms of sin and cos. Get sine squared, get cos squared from that. Boom you got a quadratic in terms of cos.
    Ok so I'm up to...

    3cos \theta = 2sin^2 \theta

    how do I get cos squared from that?
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    (Original post by Philip-flop)
    Ok so I'm up to...

    3cos \theta = 2sin^2 \theta

    how do I get cos squared from that?
    \sin^2 \theta+\cos^2 \theta \equiv 1
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    (Original post by RDKGames)
    \sin^2 \theta+\cos^2 \theta \equiv 1
    I'm not sure I can see where to use that trig identity for 3cos \theta=2sin^2 \theta

    Edit: wait yeah I can see it!! I was being silly
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    (Original post by RDKGames)
    \sin^2 \theta+\cos^2 \theta \equiv 1
    Thanks again!! I just needed prompting for that equation

    I managed to get...

    3cos \theta=2sin^2 \theta

    3cos \theta=2(1-cos^2 \theta)

    3cos \theta=2-2cos^2 \theta

    2cos^2 \theta +3cos \theta -2 =0

    (2cos \theta -1)(cos \theta+2) = 0

    Seriously appreciate your help. Sorry I'm such a pest
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    (Original post by Philip-flop)
    Thanks again!! I just needed prompting for that equation

    I managed to get...

    3cos \theta=2sin^2 \theta

    3cos \theta=2(1-cos^2 \theta)

    3cos \theta=2-2cos^2 \theta

    2cos^2 \theta +3cos \theta -2 =0

    (2cos \theta -1)(cos \theta+2) = 0

    Seriously appreciate your help. Sorry I'm such a pest
    Yeah that's correct, not a pest . So now

    2\cos \theta - 1 = 0

    or

    \cos \theta + 2 = 0
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    2 \cot^2 \theta - \cot \theta - 5=0

    ^Do I start off by using the quadratic formula on this?

    Or shall I use the Trig rule ... \cot \theta = \frac{1}{tan} =\frac{cos \theta}{sin \theta}
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    (Original post by Philip-flop)
    2 \cot^2 \theta - \cot \theta - 5=0

    ^Do I start off by factorising this?
    Yeah. Or completing the square.
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    (Original post by RDKGames)
    Yeah. Or completing the square.
    Or shall I use the Trig rule ... \cot \theta = \frac{1}{tan} =\frac{cos \theta}{sin \theta} ??
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    (Original post by Philip-flop)
    Or shall I use the Trig rule ... \cot \theta = \frac{1}{tan} =\frac{cos \theta}{sin \theta} ??
    No need, the quadratic is all in terms of a single unknown variable and that's as simple as it gets.
 
 
 
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