Solving (-1)vcos θ Identity Problem - Hi Friends!

[nesta]

Hi friends,

I am not able to understand how the below shown identity becomes (-1) power v cosθ.

cos(vπ − θ ) = cos vπ cos θ + sin vπ sin θ = (−1)power v cos θ


==> (-1)^vcos θ

Please help me understand this basic problem.

Thanks,
Nesta

 

[Dick]

You must mean v is an integer. So v=...-2,-1,0,1,2... Start working out sin(1*pi), sin(2*pi), sin(3*pi) ... and cos(1*pi), cos(2*pi), cos(3*pi) ... Do you see a pattern?

 

[nesta]

You must mean v is an integer. So v=...-2,-1,0,1,2... Start working out sin(1*pi), sin(2*pi), sin(3*pi) ... and cos(1*pi), cos(2*pi), cos(3*pi) ... Do you see a pattern?

Thank you very much, I understand now.

cos vπ cos θ + sin vπ sin θ

Since sin n*pi is always zero, we can omit this.

when,
v=0 : cos(0π)cosθ + sin0 = cosθ
v=1 : cos(1π)cosθ + 0 = -1 cosθ
v=2 : cos(2π)cosθ + 0 = 1 cosθ
v=3 : cos(3π)cosθ + 0 = -1 cosθ

so in general we can write it as (-1)to power v cos θ

Thank you very much.