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Stationary points of y=sinx+cosx 
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#1
Feb810, 04:46 PM

P: 14

1. The problem statement, all variables and given/known data
Solve the stationary points of y=sinx+cosx for domain pi<x<pi 2. Relevant equations 3. The attempt at a solution Differentiate: d/dx=cosx+sinx But how do i solve? 


#2
Feb810, 04:50 PM

P: 104

cosx+sinx=0
cosx=sinx 1=tanx?? 


#3
Feb810, 05:01 PM

P: 14

ive got the answers as.. (pi/4,rt2) and (3pi/4, rt2) Are the answers wrong?
I dont know how they got these?? because surly the x coordinate is 0??? 


#4
Feb810, 05:06 PM

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Stationary points of y=sinx+cosx
If you meant dy/dx = ..., you have made a mistake. Try again. Also, your notation is not correct. d/dx is an operator that is written to the left of some function. In contrast, dy/dx is the derivative of y with respect to x. 


#5
Feb810, 05:09 PM

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#6
Feb810, 05:10 PM

P: 14

so therefore:
dy/dx=cosx+sinx. stationary points when diff = 0 so cosx+sinx=0 where do i go from here?? 


#7
Feb810, 05:11 PM

P: 14




#8
Feb810, 07:25 PM

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#9
Feb810, 07:28 PM

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P: 21,215

To find the stationary points, set dy/dx to zero. cosx  sinx = 0 ==> cosx + sinx = 0 ==> 1 + tanx = 0 (dividing both sides by cosx) Can you continue? 


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