Question

How do you find the 6 trigonometric functions for 16pi/3?

Answer 1

`sin((16pi)/3)=-sqrt(3)/2`
`cos((16pi)/3)=-1/2`
`tan ((16pi)/3)=sqrt(3)`
`ctg ((16pi)/3)=sqrt(3)/3`
`sec((16pi)/3)=-2`
`csc((16pi)/3)=-(2sqrt(3))/2`

Explanation:

First you have to notice, that the angle is greater than `2pi`, so you have to reduce it first.

`(16pi)/3=5 1/3 pi=4pi+(4pi)/3`, so for all functions you can calculate them for `(4pi)/3` instead of the original value.

Now you can start calculating the values of trig functions:

`sin((4pi)/3)=sin(pi+pi/3)=-sin(pi/3)=-sqrt(3)/2`
`cos((4pi)/3)=cos(pi+pi/3)=-cos(pi/3)=-1/2`

To calculate these values I used 2 facts:

1. `(4pi)/3` is in the 3rd quadrant, so tangent and cotangent are positive, other functions are negative
2. If you have an even multiple of `pi/2` in the reduction formula, function does not change to a co-function.

Now we can use the trigonometric identities to calculate other 4 functions:

`tan ((4pi)/3)=(sin((4pi)/3))/(cos((4pi)/3))=sqrt(3)`

`ctg ((4pi)/3)=1/tan((4pi)/3)=sqrt(3)/3`

`sec ((4pi)/3)=1/cos ((4pi)/3) = -2`

`csc ((4pi)/3)=1/ sin ((4pi)/3)=-(2sqrt(3))/3`