# `cos(3pi)` Evalute the trigonometric function using its period as an aid

*print*Print*list*Cite

### 1 Answer

`cos(3pi)`

the periodicity of `cos (x)` is `cos(x+2pi)`

so,

`cos(3pi)` = `cos(pi+2Pi)`

= `cos(pi) = -1`

so,

`cos(3pi) = -1`

*print*Print*list*Cite

`cos(3pi)`

the periodicity of `cos (x)` is `cos(x+2pi)`

so,

`cos(3pi)` = `cos(pi+2Pi)`

= `cos(pi) = -1`

so,

`cos(3pi) = -1`

We have a Math tutor online right now to help you!

To chat with a tutor, please set up a tutoring profile by creating an account and setting up a payment method. If you initiate a chat, please note you will be charged $0.50 a minute for tutoring time.