C3 Trigonometry

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  1. TLK's Avatar
    21 22-03-2012  20:08
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    Re: C3 Trigonometry
    (Original post by hassi94)
    That's because it's trying to do tan(90) (undefined) then doing 2/it, instead of just doing 2cos(90) then dividing by sin90.
    I see what you did there:adore:
  2. GreenLantern1's Avatar
    22 22-03-2012  20:20
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    Re: C3 Trigonometry
    (Original post by hassi94)
    210 is, but it seems you've used an incorrect method (just got one correct answer coincidentally).

    Right so if I ask you to write down the values of 3x/2 for the range 0<x<360 what are your answers?
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
  3. shake_it's Avatar
    23 22-03-2012  20:26
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    Re: C3 Trigonometry
    tan(3x/2)=-1
    a=3x/2
    tan(a)=-1 => a=3(pi)/4 + k(pi), k is an integer.
    3x/2=3(pi)/4 + k(pi)
    3x=3(pi)/2 + 2k(pi)
    x=(pi)/2 +2k(pi)/3, k is an integer.
  4. Intriguing Alias's Avatar
    24 22-03-2012  20:59
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    Re: C3 Trigonometry
    (Original post by GreenLantern1)
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
  5. GreenLantern1's Avatar
    25 22-03-2012  21:16
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    Re: C3 Trigonometry
    (Original post by hassi94)
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
  6. Intriguing Alias's Avatar
    26 22-03-2012  21:17
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    Re: C3 Trigonometry
    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Yep.

    Now do you still need help with that other question?
  7. TSong's Avatar
    27 22-03-2012  21:19
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    Re: C3 Trigonometry
    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Do you know the general solution for tangent?

    If you consider the graph of tanx, and x = 0.5,
    http://www.wolframalpha.com/input/?i=plot+tanx+and+0.5
    they meet once in every period.
    Last edited by TSong; 22-03-2012 at 21:34.
  8. GreenLantern1's Avatar
    28 22-03-2012  21:46
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    Re: C3 Trigonometry
    (Original post by hassi94)
    Yep.

    Now do you still need help with that other question?
    Yes please

    I am just clueless as to how to approach it!
  9. Intriguing Alias's Avatar
    29 22-03-2012  22:17
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    Re: C3 Trigonometry
    (Original post by GreenLantern1)
    Yes please

    I am just clueless as to how to approach it!
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

    \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
  10. GreenLantern1's Avatar
    30 22-03-2012  22:30
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    Re: C3 Trigonometry
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

    \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    So the angles in answer to your question are x= 0,180,360,66.4,293.6
  11. GreenLantern1's Avatar
    31 22-03-2012  22:42
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    Re: C3 Trigonometry
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

    \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
  12. GreenLantern1's Avatar
    32 22-03-2012  22:45
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    Re: C3 Trigonometry
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

    \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    I also did another question: 2sin^2x=3cosx and i got values of 60,300 after making it into a quadratic. Is that correct; I think it is?
  13. Intriguing Alias's Avatar
    33 22-03-2012  23:00
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    Re: C3 Trigonometry
    (Original post by GreenLantern1)
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
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  14. GreenLantern1's Avatar
    34 22-03-2012  23:02
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    Re: C3 Trigonometry
    (Original post by hassi94)
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
    Thanks, I was just a bit stuck at first
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