Question

How do you find the six trigonometric functions of `(13pi)/4`?

Answer 1

Six trig functions of `(13pi)/4`

Explanation:

On the trig unit circle,

`sin ((13pi)/4) = sin (5pi/4 + 2pi)` = `- sin (pi/4) = (-sqrt2)/2`
`cos ((13pi)/4) = cos (5pi/4 + 2pi) = -cos (pi/4) = (-sqrt2)/2`

`tan ((13pi)/4) = 1`

`cot ((13pi)/4) = 1`

`sec ((13pi)/4) = -sqrt2`

`csc ((13pi)/4) = -sqrt2`