So the statement sin(n+1)pi/2 = cos(n)pi/2 is true by this property.

[SVXX]

Question - Find the Fourier series for f(x) = |cos(x)| in the interval (-π, π).
Right, I attempted the question and the integration that followed. I'm having trouble in the integration itself..
Firstly I found that f(x) is an even function, so the sine term(b_n) of the Fourier expansion would be zero and that I can use a property of integration to write

as

and similarly so for a_n.

I computed a₀ to be zero, but the trouble is, I'm getting a_n to be zero too, and if that is so, there will be no Fourier series! Could someone please point out my mistake?

 

[uart]

Hints :

1. One period of |cos(x)| extends from -pi/2 to pi/2. Check your limits.

2. Beware divide by zero.

 

[SVXX]

Thanks for the hints...I figured out what I was doing wrong in the integration. I had to break the intervals at pi/2 as |cos(x)| = cos(x) for 0 < x < pi/2 and = - cos(x) for pi/2 < x < pi.
I referred a popular book from the college library today which proved me correct. But what I don't understand is the last part of the solution that was given in that book...it stated that sin(n+1)pi/2 = cos(n)pi/2...how is this so?

 

[uart]

Thanks for the hints...I figured out what I was doing wrong in the integration. I had to break the intervals at pi/2 as |cos(x)| = cos(x) for 0 < x < pi/2 and = - cos(x) for pi/2 < x < pi.
I referred a popular book from the college library today which proved me correct. But what I don't understand is the last part of the solution that was given in that book...it stated that sin(n+1)pi/2 = cos(n)pi/2...how is this so?

The complementary relation between sin and cos tells us that sin x = cos (pi/2 - x) and visa versa.

So sin (n+1)pi/2 = cos (pi/2 -(n+1)pi/2) = cos (-n pi/2).

Now since cos is an even function then cos (-n pi/2) = cos (n pi/2).