C3 Trig Help

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  1. 0range's Avatar
    1 13-06-2012  18:15
    • Exalted and Worshipped Member
    C3 Trig Help


    I've done part A, but a bit confused on part B

    I know the Minimum is -5, but shouldn't this occur when sin(x+0.6435) = -1

    On the mark scheme it occurs when sin(x+0.6435) = 3(pi)/2
  2. raheem94's Avatar
    2 13-06-2012  18:26
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    • Posts: 5,512
    Re: C3 Trig Help
    (Original post by 0range)


    I've done part A, but a bit confused on part B

    I know the Minimum is -5, but shouldn't this occur when sin(x+0.6435) = -1

    On the mark scheme it occurs when sin(x+0.6435) = 3(pi)/2
    It should be x + 0.6435 = \dfrac{3 \pi}2

    Can you give the link to the mark scheme.
  3. 0range's Avatar
    3 13-06-2012  18:36
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    Re: C3 Trig Help
    (Original post by raheem94)
    It should be x + 0.6435 = \dfrac{3 \pi}2

    Can you give the link to the mark scheme.
    It is x + 0.6435 = \dfrac{3 \pi}2 my bad
    It's downloaded so I'll try n upload it, but it's Soloman Paper C Q7
    Attached Files
  4. File Type: pdf C3Cmarks.pdf (162.1 KB, 57 views)
    5. Last edited by 0range; 13-06-2012 at 18:38.
  5. raheem94's Avatar
    4 13-06-2012  18:41
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    Re: C3 Trig Help
    (Original post by 0range)
    It is x + 0.6435 = \dfrac{3 \pi}2 my bad
    It's downloaded so I'll try n upload it, but it's Soloman Paper C Q7
    \displaystyle \sin (x + 0.6435) = -1 \implies x + 0.6435 = \sin^{-1}(-1) = \dfrac{3\pi}{2} \\ \implies x = \dfrac{3 \pi}2 - 0.6435 \implies \boxed{x = 4.07 }

    Do you get it?
  6. 0range's Avatar
    5 13-06-2012  18:43
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    Re: C3 Trig Help
    (Original post by raheem94)
    \displaystyle \sin (x + 0.6435) = -1 \implies x + 0.6435 = \sin^{-1}(-1) = \dfrac{3\pi}{2} \\ \implies x = \dfrac{3 \pi}2 - 0.6435 \implies \boxed{x = 4.07 }

    Do you get it?
    Yh I did almost exactly that, but when I put Sin^-1(-1) I get -0.5(pi) :s
  7. raheem94's Avatar
    6 13-06-2012  18:55
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    Re: C3 Trig Help
    (Original post by 0range)
    Yh I did almost exactly that, but when I put Sin^-1(-1) I get -0.5(pi) :s
    We need the smallest positive value of x .

    \displaystyle \sin^{-1}(-1) = - \frac{\pi}2 , \ \frac{3 \pi}2, \ldots

    We need the positive value hence we will use the other solution. The solution you used will give a negative value of x

    Does it makes sense?
  8. 0range's Avatar
    7 13-06-2012  19:01
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    Re: C3 Trig Help
    (Original post by raheem94)
    We need the smallest positive value of x .

    \displaystyle \sin^{-1}(-1) = - \frac{\pi}2 , \ \frac{3 \pi}2, \ldots

    We need the positive value hence we will use the other solution. The solution you used will give a negative value of x

    Does it makes sense?

    Yh it makes sense Thank you


    Could you help me with one more thing?

    For Rconversion I use Phythagerous for R and for alpha I use tan^-1(b/a)

    Would there ever be a case when to work out alpha you would have to do tan-1(a/b)? or tan^-1(-b/a)
    Sorry I know it's really obscure but I haven't come across anything like that in any of the past papers. Thanks again!
  9. raheem94's Avatar
    8 13-06-2012  19:11
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    Re: C3 Trig Help
    (Original post by 0range)
    Yh it makes sense Thank you


    Could you help me with one more thing?

    For Rconversion I use Phythagerous for R and for alpha I use tan^-1(b/a)

    Would there ever be a case when to work out alpha you would have to do tan-1(a/b)? or tan^-1(-b/a)
    Sorry I know it's really obscure but I haven't come across anything like that in any of the past papers. Thanks again!
    I don't generalize formulae, i do it by looking at the question.

    I will approach these in the following way,

    R \cos \alpha = 4 \longrightarrow (1) \\ R \sin \alpha = 3 \longrightarrow (2)

    Now square both the above equations.

    R^2 \cos^2 \alpha = 16 \longrightarrow (3) \\ R^2 \sin^2 \alpha = 9 \longrightarrow (4)

    Now add (3) \text{ and } (4)

    R^2 \cos^2 \alpha+ R^2 \sin^2 \alpha = 16+9 \implies R^2 ( \sin^2 \alpha + \cos^2 \alpha ) = 25 \\ \implies R^2 (1) = 25 \implies R^2 = 25 \implies \boxed{ R = 5 }

    Now to find \alpha , do (2) \div (1) .
    \displaystyle \frac{R \sin \alpha }{R \cos \alpha} = \frac34 \implies \tan \alpha = \frac34

    Do you get it?
  10. Bugsy's Avatar
    9 13-06-2012  19:15
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    • Posts: 158
    Re: C3 Trig Help
    (Original post by raheem94)
    We need the smallest positive value of x .

    \displaystyle \sin^{-1}(-1) = - \frac{\pi}2 , \ \frac{3 \pi}2, \ldots

    We need the positive value hence we will use the other solution. The solution you used will give a negative value of x

    Does it makes sense?
    Hi sorry to intrude on the thread but I just had a quick question about these types of questions. When you want to express something in this form, once I find out the value of R I don't use tan to work out the alpha value. I use another method where I just compare what is similar in the orginal equation and what's not. Like, in the example that's on this thread, I would expand out R sin (x+ alpha) into R sin x cos alpha + R cos x sin alpha. Then with the first half, I would see that sin x is present in both equations so I would use R cos alpha = 4 and then solve it from there using cos-1 etc. I've not seen this quoted in the mark schemes but it always gives me the correct answer, would this method be alright to use :confused: Thanks so much!
  11. raheem94's Avatar
    10 13-06-2012  19:16
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    Re: C3 Trig Help
    (Original post by Bugsy)
    Hi sorry to intrude on the thread but I just had a quick question about these types of questions. When you want to express something in this form, once I find out the value of R I don't use tan to work out the alpha value. I use another method where I just compare what is similar in the orginal equation and what's not. Like, in the example that's on this thread, I would expand out R sin (x+ alpha) into R sin x cos alpha + R cos x sin alpha. Then with the first half, I would see that sin x is present in both equations so I would use R cos alpha = 4 and then solve it from there using cos-1 etc. I've not seen this quoted in the mark schemes but it always gives me the correct answer, would this method be alright to use :confused: Thanks so much!
    Both methods are fine
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  12. 0range's Avatar
    11 13-06-2012  19:17
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    Re: C3 Trig Help
    (Original post by raheem94)
    I don't generalize formulae, i do it by looking at the question.

    I will approach these in the following way,

    R \cos \alpha = 4 \longrightarrow (1) \\ R \sin \alpha = 3 \longrightarrow (2)

    Now square both the above equations.

    R^2 \cos^2 \alpha = 16 \longrightarrow (3) \\ R^2 \sin^2 \alpha = 9 \longrightarrow (4)

    Now add (3) \text{ and } (4)

    R^2 \cos^2 \alpha+ R^2 \sin^2 \alpha = 16+9 \implies R^2 ( \sin^2 \alpha + \cos^2 \alpha ) = 25 \\ \implies R^2 (1) = 25 \implies R^2 = 25 \implies \boxed{ R = 5 }

    Now to find \alpha , do (2) \div (1) .
    \displaystyle \frac{R \sin \alpha }{R \cos \alpha} = \frac34 \implies \tan \alpha = \frac34

    Do you get it?
    So imagine if it was 3sinx - 4cosx

    then Rcosx = 3
    Rsinx = -4

    therefore it would be tan^-1(-4/3) ?
  13. Bugsy's Avatar
    12 13-06-2012  19:20
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    • Posts: 158
    Re: C3 Trig Help
    (Original post by raheem94)
    Both methods are fine
    Thank you!!
  14. raheem94's Avatar
    13 13-06-2012  19:26
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    Re: C3 Trig Help
    (Original post by 0range)
    So imagine if it was 3sinx - 4cosx

    then Rcosx = 3
    Rsinx = -4

    therefore it would be tan^-1(-4/3) ?
    Express it as -3 \sin (x) - 4 \cos (x) \equiv R \sin ( x - \alpha)

    For such questions 0 < \alpha < \dfrac{\pi}2 . so we won't use your value.

    Use my method.

    Though if it is done by your method then also it can be made positive in the following way,
    \displaystyle R \sin \left( x + \tan^{-1} \left( - \frac43 \right) \right) = R \sin \left( x - \tan^{-1} \left( \frac43 \right) \right)
  15. 0range's Avatar
    14 13-06-2012  19:29
    • Exalted and Worshipped Member
    Re: C3 Trig Help
    (Original post by raheem94)
    Express it as -3 \sin (x) - 4 \cos (x) \equiv R \sin ( x - \alpha)

    For such questions 0 < \alpha < \dfrac{\pi}2 . so we won't use your value.

    Use my method.

    Though if it is done by your method then also it can be made positive in the following way,
    \displaystyle R \sin \left( x + \tan^{-1} \left( - \frac43 \right) \right) = R \sin \left( x - \tan^{-1} \left( \frac43 \right) \right)
    Thank you!!! that was really helpful I'll +rep as soon as I'm able to
  16. raheem94's Avatar
    15 13-06-2012  19:31
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    Re: C3 Trig Help
    (Original post by 0range)
    Thank you!!! that was really helpful I'll +rep as soon as I'm able to
    No problem, you are welcome.
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