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

Question

anonymous · Answer

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

anonymous · Answer

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

zehanz · Answer

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

anonymous · Answer

still slightly confused

zehanz · Answer

This does give you a start, btw

anonymous · Answer

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

zehanz · Answer

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

anonymous · Answer

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

zehanz · Answer

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.

anonymous · Answer

alright thanks for your help!

anonymous · Answer

I think I know what I did wrong thank you

zehanz · Answer

Maybe it would be a good idea to use the formula of Simpson here:

$$\cos(x)+\cos(y)=2\cos \frac{ x+y }{ 2 }\cos \frac{ x-y }{ 2 }$$

zehanz · Answer

YW!