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# Trig Question watch

1. 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?

thanks
2. (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?

thanks
Why?
3. (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
4. because when cosx= 0 the graph crosses at 90 and 270?
5. (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!
6. Concerning the question:

You can't just expand the LHS,

Instead inverse cos both sides to have:

7. yeah i got that so x= 0
but then isn't it 0, 90 and 360??
8. (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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Updated: October 1, 2010
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