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    i need help with b)ii) if someone could do the working it would be really helpful being trying at it for an hour and stil cant figure it out. i checked the answers also but no clue.
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    (Original post by djmans)
    i need help with b)ii) if someone could do the working it would be really helpful being trying at it for an hour and stil cant figure it out. i checked the answers also but no clue.
    It's inverse cos of 0 first isn't it?
    so
    cos(2x-30)=0

2x-30=90

cast\ diagram\

2x-30=90,270,450

2x=120,300,480

x=60,150,240
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    (Original post by will'o'wisp)
    It's inverse cos of 0 first isn't it?
    so
    cos(2x-30)=0

2x-30=90

cast\ diagram\

2x-30=90,270,450

2x=120,300,480

x=60,150,240
    thats easy but how do you do b)ii) finding p and q
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    (Original post by djmans)
    thats easy but how do you do b)ii) finding p and q
    Oh right u didn't specify, nvm

    So this is just graph transformations isn't it?

    well we'll do q first because it's easier

    so we know with that a cos graph it touches the x axis at 90°

    A is the first bit of the graph we see from the diagram which touches the x axis at 100 degrees

    There's a difference of 10 between them

    Remember if we have y=f(x-a) we move the graph a units to the right

    so q then is simply just 10
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    (Original post by will'o'wisp)
    Oh right u didn't specify, nvm

    So this is just graph transformations isn't it?

    well we'll do q first because it's easier

    so we know with that a cos graph it touches the x axis at 90°

    A is the first bit of the graph we see from the diagram which touches the x axis at 100 degrees

    There's a difference of 10 between them

    Remember if we have y=f(x-a) we move the graph a units to the right

    so q then is simply just 10
    the answer says q=60
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    (Original post by djmans)
    thats easy but how do you do b)ii) finding p and q
    p is a bit more difficult

    so we know the difference between where a cos graph normally touches the x axis is at the points 90° and 270° between the graph drawn in the range 0 to 360

    the difference between A and B is 120

    yet the original cos graph has a difference of 180 so to get the ratio between the given graph and the cos graph we do 120/180 which is 2/3

    remember y=f(2x) we divide the x value by 2 not multiply the x value by 2 so in a sense we do 1/2

    so for our 2/3 instead of 2/3 we do \dfrac{1}{\frac{2}{3}}

    which is of course 3/2
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    (Original post by will'o'wisp)
    p is a bit more difficult

    so we know the difference between where a cos graph normally touches the x axis is at the points 90° and 270° between the graph drawn in the range 0 to 360

    the difference between A and B is 120

    yet the original cos graph has a difference of 180 so to get the ratio between the given graph and the cos graph we do 120/180 which is 2/3

    remember y=f(2x) we divide the x value by 2 not multiply the x value by 2 so in a sense we do 1/2

    so for our 2/3 instead of 2/3 we do \dfrac{1}{\frac{2}{3}}

    which is of course 3/2
    p is right but q is supposed to 60
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    (Original post by djmans)
    the answer says q=60
    (Original post by djmans)
    p is right but q is supposed to 60
    hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

    plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them :confused::confused:
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    (Original post by djmans)
    the answer says q=60
    \cos(100p-q)=0 \Rightarrow 100p-q=90
    \cos(220p-q)=0 \Rightarrow 220p-q=270

    from considering cosine's periodicity and the interval in which A and B are in.

    A different way to think about it would be to take \cos(x)=0 and you know that this crosses at x=90 and x=270
    Now do the mapping x \mapsto px-q and you have px-q=90 and px-q=270 then just substitute the appropriate angles in.
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    (Original post by will'o'wisp)
    hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

    plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them :confused::confused:
    hmm could it be the answer in the book is wrong after all this question is from the text book.
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    (Original post by RDKGames)
    \cos(100p-q)=0 \Rightarrow 100p-q=90
    \cos(220p-q)=0 \Rightarrow 220p-q=270

    from considering cosine's periodicity and the interval in which A and B are in.
    so we supposed to add the differences together? o.o
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    (Original post by will'o'wisp)
    hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

    plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them :confused::confused:
    The problem with your solution is that you're trying to work out q without taking how stretched the graph is into account. RDKGames's solution avoids this problem.
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    (Original post by djmans)
    hmm could it be the answer in the book is wrong after all this question is from the text book.
    sorry, since i forgot to take into account stretchiness of graph, multiply 100° by 1.5 which is p then we can see the difference because multiplying by 1.5 removes the stretchiness and makes the graph "normal", we then see A is 150° and the difference between 90 and 150 is 60
    (Original post by TimGB)
    The problem with your solution is that you're trying to work out q without taking how stretched the graph is into account. RDKGames's solution avoids this problem.
 
 
 
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