Write the principal value of `cos^(-1)[cos(680^0)]dot`

Explanation

Principal value of `cos^-1(cos680^@)` will lie between `0` and `180^@`.


Now, `cos(680^@) = cos(360^@+320^@) = cos320^@`

`=cos(360^@-40^@) = cos(-40^@) = cos40^@` ....(As `cos(-theta) = costheta`)

`:. cos^-1[cos680^@] = cos^-1[cos40^@] = 40^@`

So, principal value of given expression is `40^@`.