Question

How to add cos(40pi*t+pi)+cos(-40*pi*t-pi)

Answer 1

I think cos(-40*pi*t-pi) becomes -cos(40*pi*t+pi)?

Answer 2

oh snap sorry I meant How to add cos(40pi*t-pi)+cos(-40*pi*t-pi)

Answer 3

No, for every x, cos(-x)=cos(x); cos (x) is an even function.

Answer 4

still slightly confused

Answer 5

This does give you a start, btw

Answer 6

one is a shift of -pi and one is a shift of +pi?

Answer 7

It means cos(-40pi*t - pi)=cos(40pi*t+pi)

Answer 8

my question was actually cos(40pi*t-pi) sorry so would cos(-40pi*t-pi) still equal that

Answer 9

Not really, you have 40pi*t, so to see the shift, you would have to write it as:

cos(40pi(t-1/40)), so the shift is only 1/40.

Answer 10

alright thanks for your help!

Answer 11

I think I know what I did wrong thank you

Answer 12

Maybe it would be a good idea to use the formula of Simpson here:

\[\cos(x)+\cos(y)=2\cos \frac{ x+y }{ 2 }\cos \frac{ x-y }{ 2 }\]

Answer 13

YW!