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    hey guys,

    i was doing a question which involved writing sin in terms of cos and long story short, i kept thinking that sinx is equal to cos(x-(pi/2)), where it is supposed to be cos ((pi?2)-x).

    I was trying to see why and when i used the cos(a-b) i did get sin x, but when looking at it graphically (clearer for me), i just dont sinx from cos ((pi?2)-x).

    i have tried to illustrate my problem in paint (see attached ) but as can be seen i keep getting -sinx as opposed to sin x

    any help is much appreciated
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    (Original post by type_writer)
    hey guys,

    i was doing a question which involved writing sin in terms of cos and long story short, i kept thinking that sinx is equal to cos(x-(pi/2)), where it is supposed to be cos ((pi?2)-x).

    I was trying to see why and when i used the cos(a-b) i did get sin x, but when looking at it graphically (clearer for me), i just dont sinx from cos ((pi?2)-x).

    i have tried to illustrate my problem in paint (see attached ) but as can be seen i keep getting -sinx as opposed to sin x

    any help is much appreciated
    Your error is due to your incorrect usage of the composition of functions.
    Write the separate functions down in terms of f(x) and g(x) and the mistake will become apparent.

    An easy way to remember and see the connection between sin and cos is to draw a right angled triangle with sides a,b and c and angle of pi/2, x and (pi/2 - x) ... A little playing around and you will see the desired results.

    (Original post by SimonM)
    They're the same thing. \cos (x) = \cos (-x)




    ***Edit
    \\simon is of course quite correct... Just to clarify, the first point I make is related to the error that you have made on the attachment with regards to the composition of functions.....
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    They're the same thing. \cos (x) = \cos (-x)
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    i knew that cosx = cos(-x) (is on the 3rd line of the attachment in the OP)

    i just thought that, like other functions, if i add pi/2 to it, the graph would move pi/2 to the left and hence the confusion as what i have drawn is cos (pi/2 -x) but its is not the same as sin x. i still dont quite get what im doing wrong here...
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    (Original post by type_writer)
    i knew that cosx = cos(-x) (is on the 3rd line of the attachment in the OP)

    i just thought that, like other functions, if i add pi/2 to it, the graph would move pi/2 to the left and hence the confusion as what i have drawn is cos (pi/2 -x) but its is not the same as sin x. i still dont quite get what im doing wrong here...
    as I mentioned, your problem lies in your application/understanding of composite functions.
    see earlier post..
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    (Original post by type_writer)
    i knew that cosx = cos(-x) (is on the 3rd line of the attachment in the OP)

    i just thought that, like other functions, if i add pi/2 to it, the graph would move pi/2 to the left
    If you want to draw the graph first take out the mius sign from the inner function
    to see wich direction have you to move the graph.
    cos(pi/2-x)=cos[-(x-pi/2)]=cos(x-pi/2)
    and hence the confusion as what i have drawn is cos (pi/2 -x) but its is not the same as sin x. i still dont quite get what im doing wrong here...
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    (Original post by ztibor)
    If you want to draw the graph first take out the mius sign from the inner function
    to see wich direction have you to move the graph.
    cos(pi/2-x)=cos[-(x-pi/2)]=cos(x-pi/2)
    Exactly that

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

    Hopefully the latex code makes it clearer to see
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    ahh right, i see it now!!!

    thanks a lot.
 
 
 
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