Simplifying Trigonometric Functions with Arbitrary n

[Deathfish]

ok i know that

sin n*(pi/2)
= 1 if n=1,5,9,13...
= -1 if n=3,7,11,15...
= 0 if n is even

cos n*(pi/2)
= 0 if n is odd
= -1 if n=0,4,8,12
= 1 if n=2,6,10,14...

is there a simpler way of expressing this?
for example simple way to express cos(n*pi)=cos(-n*pi)=(-1)^n

is there a similar way to express cos n*(pi/2) and sin n*(pi/2)
thanks

 

[tiny-tim]

Hi Deathfish!

is there a similar way to express cos n*(pi/2) and sin n*(pi/2)

standard trick …

sin((2n + 1)π/2) = (-1)ⁿ

and i suppose

cos((2n)π/2) = (-1)ⁿ ​

 

[Deathfish]

and how do you use the term in Fourier series?

lets say you encounter the term .. sin n*(pi/2)

just replace with sin((2n + 1)π/2) ? don't know how to use this expression properly

any simple example will be helpful.

 

[tiny-tim]

i'm not seeing what the difficulty is …

you just replace the sin, or cos, with (-1)^something ​

 

[Deathfish]

ok what is the (something)

 

[tiny-tim]

sin((2n + 1)π/2) = (-1)ⁿ



cos((2n)π/2) = (-1)ⁿ ​

 

[Deathfish]

ok is it because the 'n' is arbitrary you can just replace sin n*(pi/2) with sin((2n + 1)π/2) ?

 

[tiny-tim]

ok is it because the 'n' is arbitrary you can just replace sin n*(pi/2) with sin((2n + 1)π/2) ?

abritrary and odd

yes

(though of course, it's a different n …

any odd n is 2m + 1, then we rename m as n )​