Core 3 Trigonometry

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  1. Pika-Profiterole's Avatar
    • Full Member
    • Posts: 107
    Core 3 Trigonometry
    Good afternoon.

    I would like some help for some questions on trigonometry in Core 3 Maths.

    There is one question that I tried many times but I did not get it right:

    Show that cosec x + cot x = 1/ cosec x - cot x (cosec x does not equal to cot x).

    In my calculation (as below):

    cosec x + cot x = 1/ cosec x - cot x

    1/ Sin x + 1/ Tan x

    1/ sin x + cos x/ sin x

    1 + cos x / sin x

    (1 + cos x / sin x)^2

    1 + cos^2 x / sin^2 x

    1 + (1-sin^2 x)/ sin^2 x

    2 - sin^2 x/ sin^2 x

    (2/sin^2 x) - 1

    2 (1/sin^2 x) -1

    2 cosec^2 x - 1 (That's the final answer I have calculated)

    So have I made a massive mistake or over complicated anything in my calculation? (I always over-complicate things in my calculation >.<`)

    Please do tell show me on which stage did I done it wrong! Hopefully I will not make the same mistake in exams never ever again!

    Many thanks,
  2. TenOfThem's Avatar
    • TSR Royalty
    Re: Core 3 Trigonometry
    Are you trying to show that

    cosec(x) + cot(x) = \frac{1}{cosec(x)-cot(x)}

    RHS

    cosec - cot = \frac{1}{sin} - \frac{cos}{sin} = \frac{1-cos}{sin}

    \frac{1}{cosec-cot} = \frac{sin}{1-cos} = \frac{sin(1+cos)}{1-cos^2}


    Can you finish this off
  3. Pika-Profiterole's Avatar
    • Full Member
    • Posts: 107
    Re: Core 3 Trigonometry
    (Original post by TenOfThem)
    Are you trying to show that

    cosec(x) + cot(x) = \frac{1}{cosec(x)-cot(x)}

    RHS

    cosec - cot = \frac{1}{sin} - \frac{cos}{sin} = \frac{1-cos}{sin}

    \frac{1}{cosec-cot} = \frac{sin}{1-cos} = \frac{sin(1+cos)}{1-cos^2}


    Can you finish this off
    @o@
    May I ask, how did you get \frac{1}{cosec-cot} = \frac{sin}{1-cos} = \frac{sin(1+cos)}{1-cos^2}?

    would you mind to explain step by step? (Sorry!! I really don't understand it >.<)
  4. TenOfThem's Avatar
    • TSR Royalty
    Re: Core 3 Trigonometry
    I turned the fraction over

    And then I multiplied numerator and denomination by (1+cos)
  5. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: Core 3 Trigonometry
    (Original post by Pika-Profiterole)
    ...
    You have overcomplicated the question a tiny bit.

    Why not just show that \mathrm{cosec} ^2 x-\cot^2 x=1?

    (it can be done in two lines)

    For future reference, Maths questions belong here
  6. Pika-Profiterole's Avatar
    • Full Member
    • Posts: 107
    Re: Core 3 Trigonometry
    Yes, I done it but in a different way.

    I multiply both side by (cosec x - cot x)-- and my final answer is:
    1 - cos^2x = sin^2

    It just so different to the method that my teacher taught us... but the good news is I understand it, finally!! Haha..

    Thank you so much for helping me.

    By the way, I am doing maths questions (again!) amd there are two questions that I am not really sure is my calculation correct!

    Would you like to check it for me please?
  7. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    Re: Core 3 Trigonometry
    (Original post by Pika-Profiterole)
    Yes, I done it but in a different way.

    I multiply both side by (cosec x - cot x)-- and my final answer is:
    1 - cos^2x = sin^2

    It just so different to the method that my teacher taught us... but the good news is I understand it, finally!! Haha..

    Thank you so much for helping me.

    By the way, I am doing maths questions (again!) amd there are two questions that I am not really sure is my calculation correct!

    Would you like to check it for me please?
    No problem, PM your working to me.
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