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    Using cosx-sinx = (square root of 2) cos ( x + 1/4 pie)

    find the exact solutions in the interval 0<x<2 pie of the equation cosx - sinx =1 '

    what i've done is..

    (square root of 2) cos (x + 1/4 pie) = 1
    then cos (x + 1/4 pie) = 1 / (square root of 2 )
    then cos x + cos 1/4 pie = 1/ square root of pie

    and then finally i get cosx = 0 and the solutions are 90 and 270 aren't they? surely?

    helpppp please
    thanks
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    (Original post by sophiestanmoor110)
    Using cosx-sinx = (square root of 2) cos ( x + 1/4 pie)

    find the exact solutions in the interval 0<x<2 pie of the equation cosx - sinx =1 '

    what i've done is..

    (square root of 2) cos (x + 1/4 pie) = 1
    then cos (x + 1/4 pie) = 1 / (square root of 2 )
    then cos x + cos 1/4 pie = 1/ square root of pie

    and then finally i get cosx = 0 and the solutions are 90 and 270 aren't they? surely?

    helpppp please
    thanks
    Why?
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    (Original post by sophiestanmoor110)
    (square root of 2) cos (x + 1/4 pie) = 1
    then cos (x + 1/4 pie) = 1 / (square root of 2 )
    then cos x + cos 1/4 pie = 1/ square root of pie
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    because when cosx= 0 the graph crosses at 90 and 270?
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    (Original post by sophiestanmoor110)
    because when cosx= 0 the graph crosses at 90 and 270?
    you need to look at the bit I put in bold.

    Also, clarity raises a very good (and tasty) point!
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    Concerning the question:

     \cos(x+ \frac{\pi}{4}) = \dfrac{1}{\sqrt{2}}

    You can't just expand the LHS,  \cos(x+\frac{\pi}{4}) \not= \cos(x) + \cos(\frac{\pi}{4})

    Instead inverse cos both sides to have:

     x + \dfrac{\pi}{4} = arccos\left(\dfrac{1}{\sqrt{2}}\  right)
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    yeah i got that so x= 0
    but then isn't it 0, 90 and 360??
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    (Original post by sophiestanmoor110)
    yeah i got that so x= 0
    but then isn't it 0, 90 and 360??
    The cosine curve does not pass through the origin. Also, it's incredible that you got the same answer using a correct and incorrect method.
 
 
 
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