how would i show that the values of x at which sinxe^(-sinx) has stationary points form an arithmetic sequence.

 

I've so far worked out

dy/dx=cosxe^-sinx(1-sinx)=0

so cosx.e^-sinx= and sinx=0

arcsin0=pi,2pi,3pi,4pi,etc

so

cospi.e^-sinpi=0

cos2pi.e^-sin2pi=1

cos3pi.e^-sin3pi=0

cos4pi.e^-sin4pi=1

 

therefore it is a repeating sequence and common difference 1,-1 but i don't think this is correct

as again (im not sure how to solve (cosxe^-sinx(1-sinx)=0))

any ideas?

thanks for any help

how would i show that the values of x at which sinxe^(-sinx) has stationary points form an arithmetic sequence.

 

I've so far worked out

dy/dx=cosxe^-sinx(1-sinx)=0

so cosx.e^-sinx= and sinx=0

arcsin0=pi' date='2pi,3pi,4pi,etc

so

cospi.e^-sinpi=0

cos2pi.e^-sin2pi=1

cos3pi.e^-sin3pi=0

cos4pi.e^-sin4pi=1[/b']

 

therefore it is a repeating sequence and common difference 1,-1 but i don't think this is correct

as again (im not sure how to solve (cosxe^-sinx(1-sinx)=0))

any ideas?

thanks for any help

 

dy = cosx(1 - sinx)e^-sinx is correct.

dx

dy = 0 => cosx = 0 or 1 - sinx = 0 or e^-sinx = 0.

dx

cosx = 0 => x = pi/2 + n*pi, where n is an integer.

sinx = 1 => cosx = 0, so this adds no new solutions.

e^-sinx is never equal to zero, as the exponential function never equals zero.

 

So the turning points occur at pi/2 + n*pi; i.e., they form an arithmetic sequence.

thanks very much for your help!