Question

Question #61a0b

Answer 1

`tan((23pi)/3)=-tan(pi/3)=-sqrt3`

Explanation:

consider the diagram of the quadrants above.

`tan((23pi)/3)`

each revolution of the quadrants is `2pi`

so we subtract multiples of `2pi` until we have a value

`0 <=theta <=2pi`

`((23pi)/3)-(6pi)/3=(17pi)/3>2pi`

`(17pi)/3-(6pi)/3=(11pi)/3>2pi`

`(11pi)/3-(6pi)/3=(5pi)/3<2pi`

we now decide which quadrant `(5pi)/3`is in

from the diagram

`(5pi)/3>(3pi)/2`

so it is in quadrant IV. where tan is negative

we required the angle with the `+x-`axis

`:.tan((23pi)/3)=-tan(2pi-((5pi)/3))`

`=-tan(pi/3)=-sqrt3.`