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Math Help - How Trig Cofunctions Relate to Transformations

  1. #1
    Xei
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    How Trig Cofunctions Relate to Transformations

    Hi, could somebody clear this up.

    cosx=sinpi/2-x

    I must be doing something wrong, because as far as I'm aware this is

    sin(-x) which should transform the sin graph by flipping it around the y axis, followed by sin((-x)+pi/2) which should shift the graph backward by pi/2. However this graph looks like cosx flipped around the x axis..?

    Alternatively I call this sin-(x-pi/2), so that's sin((x)-pi/2) which shifts the graph to the right by 1/4 of a period, then sin-((x)-pi/2) which flips it around the y axis, although by this time the graph is symmetrical about the axis anyway and so, again, I end up with cos(-x), cosx flipped around the x axis.

    So in conclusion by transforming I'm finding that -cosx=sinpi/2-x, which is wrong. What's up?

    Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Xei View Post
    Hi, could somebody clear this up.

    cosx=sinpi/2-x

    I must be doing something wrong, because as far as I'm aware this is

    sin(-x) which should transform the sin graph by flipping it around the y axis, followed by sin((-x)+pi/2) which should shift the graph backward by pi/2. However this graph looks like cosx flipped around the x axis..?

    Alternatively I call this sin-(x-pi/2), so that's sin((x)-pi/2) which shifts the graph to the right by 1/4 of a period, then sin-((x)-pi/2) which flips it around the y axis, although by this time the graph is symmetrical about the axis anyway and so, again, I end up with cos(-x), cosx flipped around the x axis.

    So in conclusion by transforming I'm finding that -cosx=sinpi/2-x, which is wrong. What's up?

    Thanks.
    The best way I can think to explain this is that if you think about it....the graphs of cos(x) and sin(x) are out of phase by \frac{\pi}{2}...therefore that is the easiest graphic way of showing the identity
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  3. #3
    Xei
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    Thanks, I see that, but then why aren't the identities expressed as

    cos(x-\frac{\pi}{2}), sin(x+\frac{\pi}{2})

    ?

    And also, what's wrong with my methodology above?
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  4. #4
    Lord of certain Rings
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    Well if you want to map a function f(x) to f(ax+b), then translating it first and then compressing gives different result from doing it the other way round.
    The right way is to translate first then, compress.
    In your case translate by \frac{\pi}{2} then flip, you will get what you want.

    What you have done is wrong because when you first flipped it and then translated it by \frac{\pi}2, you have essential translated in the flipped axis. In other words, from the original axis point of view, you have translated the graph by \frac{\pi}2 in the opposite direction. That is to say, you flipped, thus that gave -x, then you translated by \frac{\pi}2, which is actually a -\frac{\pi}2 in the original axis.Thus it gave you \sin (-x -\frac{\pi}2) = -\cos x
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  5. #5
    Xei
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    Hmm... why do you do the translation first?

    From how I understand functions, say we have

    sin(x)

    Then we translate to the left by pi/2 to get

    sin(x+pi/2)

    Which looks like the cosine graph.

    Now say u=x+pi/2

    So the function is currently sinu

    Then we do the flip around the axis as you say, giving

    sin-u

    Which is

    sin-(x+pi/2)

    Which is

    sin(-x-pi/2)?

    Which is incorrect...

    I don't understand why you can't say u=x-pi/2 and then the function =sin-u=sin(pi/2-x) which would achieved by translating right then flipping?

    I'm confused.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Xei View Post
    Hmm... why do you do the translation first?

    From how I understand functions, say we have

    sin(x)

    Then we translate to the left by pi/2 to get

    sin(x+pi/2)

    Which looks like the cosine graph.

    Now say u=x+pi/2

    So the function is currently sinu

    Then we do the flip around the axis as you say, giving

    sin-u

    Which is

    sin-(x+pi/2)

    Which is

    sin(-x-pi/2)?

    Which is incorrect...

    I don't understand why you can't say u=x-pi/2 and then the function =sin-u=sin(pi/2-x) which would achieved by translating right then flipping?

    I'm confused.
    Because you would get the negative sine curve...for reinforement consider the fact that the sine curve is odd...therefore sin(-x)=-sin(x)
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  7. #7
    Lord of certain Rings
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    Hey, you are flipping the wrong way. Replacing x by -x is flipping about y axis.
    The problem is you are seeing sin(-x) = -sin x and thinking about flipping the curve about x-axis. You should not do this. In the procedure explained, when you translate by pi/2, you no longer have a curve that looks like sin(In fact it is cos). Thus "sin(-x) = -sin x" will not hold.


    Quote Originally Posted by Xei View Post

    Which is

    sin-(x+pi/2) ====> Wrong

    The graphical method described actually is asking you to replace all x by -x. Thats what flipping does anyway. Thus at this step you have x + pi/2 changing to -x+pi/2

    Which is

    sin(-x-pi/2)? No,this is sin(-x+pi/2)
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