Question

The terminal side of `theta` in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of `theta`?

Answer 1

See explanation.

Explanation:

If the point in the angle's terminal side is `P=(x,y)` then the trigonometric functions can be calculated as:

`sin alpha=y/r`

`cos alpha=x/r`

`tan alpha=y/x`

`cot alpha=x/y`

`sec alpha=r/x`

`csc alpha=r/y`

where `r=sqrt(x^2+y^2)`

For the given point we have:

`r=sqrt((-6)^2+8^2)=sqrt(36+64)=10`

So the functions are:

`sin alpha=8/10=4/5`

`cos alpha=(-6)/10=-3/5`

`tan alpha=8/(-6)=-4/3`

`cot alpha=(-6)/8=-3/4`

`sec alpha=10/(-6)=-5/3=-1 2/3`

`csc alpha=10/8=5/4=1 1/4`