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Trig- Sine graph translations question help (C2 A level)

  1. May 7, 2014 #1
    Please see question 9ii:
    Paper: http://www.edexcel.com/migrationdoc...00/January 2012 - QP/6664_01_que_20120307.pdf
    Mark scheme: http://www.edexcel.com/migrationdoc...00/January 2012 - MS/6664_01_msc_20120123.pdf

    This is my working:
    First I drew the sine graph, and then worked out that P had been translated/stretched from co ordinates (0,0) to (pi/10,0), Q from (pi,0) to (3pi/10,0) and R from (2pi,0) to (11pi,0).

    Let m=ax-b
    Sub x=0, m=pi/10 (where 0 is the x co ordinate before and pi/10 is the co ordinate after) so pi/10=-b so b=pi/10.

    Sub x=pi, m=3pi/5 so 3pi/5=pi*a so a=1/2

    Hence, y=sin(0.5x+pi/10) which seems to work for R as well--is this all incorrect and why?
  2. jcsd
  3. May 7, 2014 #2


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    You have it backwards. From the graph we have that [itex]\sin (ax - b) = \sin m[/itex] is zero when [itex]x = \pi/10[/itex]. That means that when [itex]x = \pi/10[/itex] we must have [itex]m = n\pi[/itex] for some integer [itex]n[/itex] (and in fact [itex]n[/itex] must be an even integer because sin(ax - b) is shown to be increasing at P, but that's beside the point). We can choose to take n = 0 for simplicity, so that when [itex]x = \pi/10[/itex] we have [itex]m = 0[/itex].
  4. May 8, 2014 #3
    I rewrote out my simultaneous equations as a(0)-b=pi/10 and a(pi)-b=3pi/5 ...however, I've been told that the 2 angles in each equation are the wrong way around- why? Because I was thinking that, for the second equation, if you apply a stretch of scale factor a to x=pi and then translate it down by b units, then it should give 3pi/5 ...?
  5. May 8, 2014 #4


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    The point is that you want to start at [itex]3\pi/5[/itex] (which is not a zero of the sine function) and do those operations in order to end up at [itex]\pi[/itex] (which is a zero). You are starting at [itex]\pi[/itex] and trying to end up at [itex]3\pi/5[/itex]!
  6. May 11, 2014 #5
    I think I understand but please can you elaborate on why you're starting from 3pi/5 and ending up at pi? (e.g. is it something to do with the function itself, or is there a proof to this?)
  7. May 11, 2014 #6


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    Consider the following problem, which I believe you know how to answer:

    "Given the graph of [itex]f(x)[/itex], sketch the graph of [itex]f(ax - b)[/itex]."

    Now instead consider this problem:

    "Given the graph of [itex]f(ax - b)[/itex], sketch the graph of [itex]f(x)[/itex]."

    I think you have read question 9(ii) as being the first, when it's closer to the second (since you're not actually asked to sketch any graphs). You can turn problems of the second type into problems of the first type by setting [itex]y = ax - b[/itex], so that [itex]x = y/a + b/a[/itex] and the problem becomes

    "Given the graph of [itex]f(y)[/itex], sketch the graph of [itex]f(y/a + b/a)[/itex]."

    which explains why your answers relate to the correct answers in the way that they do.

    But perhaps the easiest way to solve this problem is not to think about translation and scaling of graphs but about composition of functions: starting from [itex]\pi/10[/itex] I want to multiply by [itex]a[/itex] and then subtract [itex]b[/itex] to end up at an even integer multiple of [itex]\pi[/itex], and starting from [itex]3\pi/5[/itex] I want to multiply by [itex]a[/itex] and then subtract [itex]b[/itex] to end up at the next integer multiple of [itex]\pi[/itex].
  8. May 12, 2014 #7
    Hmm..I think I understand thanks- I don't suppose you help me with another question which is similar to this please (question 4b):
    http://www.skinners-maths.co.uk/specimen A level papers/EC3paper/EC3sh_H.pdf

    I wrote out that (1)(a+b)=1 and (-5)(a+b)=-1 but that doesn't seem to work? I know you can solve it directly by substituting in the co ordinates from the graph, but how would you use this idea of graph transformations on this question?
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